Family of Circles: Complete JEE Notes

Ritesh Raj · 15 Aug 2026

30 min read12 sectionsConcept

A circle is fixed by three numbers. Write it as

x^2+y^2+2gx+2fy+c=0

and you are looking at three unknowns $g$ , $f$ and $c$ . So three independent conditions pin down exactly one circle, and any two conditions leave a whole one-parameter collection of circles behind. That collection is what this chapter is about.

The wasteful way to handle two conditions is to set up $g$ , $f$ , $c$ as unknowns, write two equations, and carry a leftover parameter through pages of algebra. The efficient way is to write down a single expression that already satisfies both conditions and carries one visible parameter $\lambda$ . Then the third condition becomes one small equation in $\lambda$ , and you are done in three lines instead of thirty.

That is the whole method. Build the family first, fit $\lambda$ last.

What you should be able to do by the end

Given any two conditions on a circle, write the family in one line without hesitating about which form to use, and then read off $\lambda$ from the third condition.

We start with lines, because the parameter trick is easier to see there and it is the same trick.

Family of lines: where the $\lambda$ trick starts

Take two lines that actually meet:

L_1\equiv a_1x+b_1y+c_1=0,\qquad L_2\equiv a_2x+b_2y+c_2=0.

Let $P$ be their point of intersection. Now look at

L_1+\lambda L_2=0.

Two things are true at once, and both matter.

It is a line. Expanding gives $(a_1+\lambda a_2)x+(b_1+\lambda b_2)y+(c_1+\lambda c_2)=0$ , which is first degree in $x$ and $y$ .

It passes through $P$ . At $P$ both $L_1$ and $L_2$ are zero, so $L_1+\lambda L_2$ is $0+\lambda\cdot 0=0$ whatever $\lambda$ is.

So $L_1+\lambda L_2=0$ is the whole pencil of lines through $P$ , with one exception: $L_2=0$ itself is never produced, because no finite $\lambda$ can kill $L_1$ entirely.

Every line of the family L1 + lambda L2 = 0 passes through the fixed point P

Situation	What to write
Line through the intersection of $L_1=0$ and $L_2=0$	$L_1+\lambda L_2=0$
A family with a parameter, asked for its fixed point	group the terms as $A+\lambda B=0$ , then solve $A=0$ , $B=0$

The second row is the same statement read backwards, and it is the version JEE prefers.

Example 1. Show that for every real $\lambda$ the line $(1+2\lambda)x+(1-\lambda)y=3+3\lambda$ passes through a fixed point, and find it.

▸ SHOW SOLUTION

Step 1: Collect the $\lambda$ terms. Bring everything to one side and group:

(x+y-3)+\lambda(2x-y-3)=0.

Step 2: Read the condition. This holds for every $\lambda$ only if both brackets vanish, because a linear expression in $\lambda$ is identically zero only when its constant term and its coefficient of $\lambda$ are both zero.

x+y-3=0\quad\text{and}\quad 2x-y-3=0.

Step 3: Solve the pair. Adding, $3x=6$ , so $x=2$ , and then $y=1$ .

Step 4: Verify. Put $(2,1)$ into the original: $(1+2\lambda)(2)+(1-\lambda)(1)=2+4\lambda+1-\lambda=3+3\lambda$ . It matches the right side for all $\lambda$ .

Answer: $\boxed{(2,1)}$

Example 2. Find the equation of the line that passes through the intersection of $3x-4y+6=0$ and $x+y+2=0$ and is perpendicular to $2x+3y=0$ .

▸ SHOW SOLUTION

Step 1: Write the family. Every line through the intersection is

(3x-4y+6)+\lambda(x+y+2)=0,

that is, $(3+\lambda)x+(\lambda-4)y+(6+2\lambda)=0$ .

Step 2: Use the perpendicularity. The given line $2x+3y=0$ has slope $-\dfrac23$ , so our line needs slope $\dfrac32$ . The family has slope $-\dfrac{3+\lambda}{\lambda-4}$ , so

-\frac{3+\lambda}{\lambda-4}=\frac32\ \Longrightarrow\ -2(3+\lambda)=3(\lambda-4)\ \Longrightarrow\ -6-2\lambda=3\lambda-12.

Step 3: Solve. $5\lambda=6$ , so $\lambda=\dfrac65$ .

Step 4: Substitute back. Multiplying the family by $5$ first keeps the arithmetic clean:

5(3x-4y+6)+6(x+y+2)=0\ \Longrightarrow\ 21x-14y+42=0\ \Longrightarrow\ 3x-2y+6=0.

Step 5: Check. The two given lines meet where $3x-4y+6=0$ and $x+y+2=0$ , which gives $x=-2$ , $y=0$ . And $3(-2)-2(0)+6=0$ , so the point is on our line. Its slope is $\dfrac32$ , perpendicular to $-\dfrac23$ .

Answer: $\boxed{3x-2y+6=0}$

Why we bother with the shortcut

Finding the intersection point first and then writing a line through it also works, and for Example 2 it is barely longer. The advantage of the family form shows up when the intersection point is ugly, or when the question never asks for it. Then you never compute it at all.

Type 1: circles through the intersection of a circle and a line

Let

S\equiv x^2+y^2+2gx+2fy+c=0,\qquad L\equiv \ell x+my+n=0,

and suppose the line cuts the circle at $A$ and $B$ . Then

S+\lambda L=0

is the family of all circles through $A$ and $B$ , with $\lambda$ any real number.

The reasoning is the same two-part check as for lines.

It is a circle. Adding $\lambda L$ changes only the $x$ , $y$ and constant terms. The coefficients of $x^2$ and $y^2$ stay at $1$ and no $xy$ term appears, which are exactly the conditions for a circle.

It passes through $A$ and $B$ . At either point $S=0$ and $L=0$ , so the whole expression is zero.

The family $S + \lambda L = 0$ , with every member through the two intersection points A and B

Written out, the member with parameter $\lambda$ is

x^2+y^2+(2g+\lambda \ell)x+(2f+\lambda m)y+(c+\lambda n)=0,

so its centre is $\left(-g-\dfrac{\lambda \ell}{2},\,-f-\dfrac{\lambda m}{2}\right)$ . As $\lambda$ runs over the reals that centre travels along a straight line, and that line is the perpendicular bisector of $AB$ . Every circle through $A$ and $B$ must have its centre there, so this is a free check on your algebra.

The mistake that costs the mark

$S+\lambda L=0$ needs $S$ written with the coefficient of $x^2$ equal to $1$ . If the question hands you $3x^2+3y^2-12x+6y-9=0$ , divide by $3$ first. Skip that step and every coefficient afterwards is off by a factor, and the $\lambda$ you compute means nothing.

Example 3. Find the circle that passes through the points where $x^2+y^2=25$ meets $3x-4y-5=0$ , and also through $(1,2)$ .

▸ SHOW SOLUTION

Step 1: Write the family.

x^2+y^2-25+\lambda(3x-4y-5)=0.

Step 2: Impose the extra point. Put $x=1$ , $y=2$ :

1+4-25+\lambda(3-8-5)=0\ \Longrightarrow\ -20-10\lambda=0.

Step 3: Solve for $\lambda$ . $\lambda=-2$ .

Step 4: Substitute back.

x^2+y^2-25-2(3x-4y-5)=0\ \Longrightarrow\ x^2+y^2-6x+8y-15=0.

Step 5: Check. At $(1,2)$ : $1+4-6+16-15=0$ . Correct. The centre is $(3,-4)$ , which does lie on the line through the origin perpendicular to $3x-4y-5=0$ , as predicted.

Answer: $\boxed{x^2+y^2-6x+8y-15=0}$

Example 4. The line $y=x$ cuts the circle $x^2+y^2-2x=0$ in the segment $AB$ . Find the circle on $AB$ as diameter.

▸ SHOW SOLUTION

Step 1: Write the family. Every circle through $A$ and $B$ is

x^2+y^2-2x+\lambda(y-x)=0,

that is $x^2+y^2-(2+\lambda)x+\lambda y=0$ .

Step 2: Turn "AB is a diameter" into a condition. A circle has $AB$ as a diameter exactly when its centre is the midpoint of $AB$ , and in particular when its centre lies on the line $AB$ . Since both $A$ and $B$ are on the family circle, the centre being on $AB$ forces $AB$ to be a diameter.

Step 3: Locate the centre. From Step 1 the centre is

\left(\frac{2+\lambda}{2},\ -\frac{\lambda}{2}\right).

Step 4: Put the centre on $y=x$ .

-\frac{\lambda}{2}=\frac{2+\lambda}{2}\ \Longrightarrow\ -\lambda=2+\lambda\ \Longrightarrow\ \lambda=-1.

Step 5: Substitute.

x^2+y^2-(2-1)x-y=0\ \Longrightarrow\ x^2+y^2-x-y=0.

Step 6: Check. Solving $y=x$ with $x^2+y^2-2x=0$ gives $2x^2-2x=0$ , so $A=(0,0)$ and $B=(1,1)$ . Their midpoint is $\left(\dfrac12,\dfrac12\right)$ , which is exactly the centre of $x^2+y^2-x-y=0$ , and the radius $\dfrac{1}{\sqrt2}$ equals half of $AB=\sqrt2$ .

Answer: $\boxed{x^2+y^2-x-y=0}$

Two conditions worth recognising instantly

For a member of the family $S+\lambda L=0$ : " $AB$ is a diameter" means the centre lies on $L$ , and "the circle is the smallest of the family" means the same thing, because the smallest circle through $A$ and $B$ is the one on $AB$ as diameter. Both give you one linear equation in $\lambda$ .

Type 2: circles through the intersection of two circles

Let

S_1\equiv x^2+y^2+2g_1x+2f_1y+c_1=0,\qquad S_2\equiv x^2+y^2+2g_2x+2f_2y+c_2=0

meet at $P$ and $Q$ . Then

S_1+\lambda S_2=0,\qquad \lambda\neq-1,

is the family of circles through $P$ and $Q$ .

Here the excluded value matters. Expanding,

(1+\lambda)(x^2+y^2)+2(g_1+\lambda g_2)x+2(f_1+\lambda f_2)y+(c_1+\lambda c_2)=0,

and dividing by $1+\lambda$ gives a genuine circle. At $\lambda=-1$ the quadratic terms cancel completely and what is left,

S_1-S_2=0,

is a straight line: the common chord $PQ$ . That is not an accident, it is the degenerate member of the family, and it is useful in its own right.

Two traps in one line

First, $S_1$ and $S_2$ must both be normalised so that the coefficient of $x^2$ is $1$ before you combine them. Second, $\lambda=-1$ is excluded from the circle family. Write $\lambda\neq-1$ every time, because examiners set questions where the algebra pushes you towards exactly that value and the correct conclusion is "no such circle exists, the locus is a line".

Example 5. Find the circle through the points common to $x^2+y^2-4x-2y-8=0$ and $x^2+y^2-2x-4y-8=0$ that also passes through $(-1,4)$ .

▸ SHOW SOLUTION

Step 1: Write the family.

(x^2+y^2-4x-2y-8)+\lambda(x^2+y^2-2x-4y-8)=0.

Step 2: Impose the point $(-1,4)$ . For the first bracket:

1+16+4-8-8=5.

For the second:

1+16+2-16-8=-5.

Step 3: Solve. So $5+\lambda(-5)=0$ , giving $\lambda=1$ .

Step 4: Substitute and divide by $1+\lambda=2$ .

2x^2+2y^2-6x-6y-16=0\ \Longrightarrow\ x^2+y^2-3x-3y-8=0.

Step 5: Check. At $(-1,4)$ : $1+16+3-12-8=0$ . Correct.

Answer: $\boxed{x^2+y^2-3x-3y-8=0}$

The same idea works when the two curves are not circles, and that is where JEE has been going lately.

Example 6 (JEE Main 2026, 23 Jan Shift 2). The points of intersection of the two conics

x^2+2y^2-6x-12y+23=0\quad\text{and}\quad 4x^2+2y^2-20x-12y+35=0

lie on a circle of radius $r$ with centre $(a,b)$ . Find $ab+18r^2$ .

▸ SHOW SOLUTION

Step 1: Use the family even though these are not circles. Any curve through all the common points is

S_1+\lambda S_2=0.

Step 2: Force it to be a circle. A second degree curve is a circle when the coefficients of $x^2$ and $y^2$ are equal and there is no $xy$ term. There is no $xy$ term here, so we only need

1+4\lambda=2+2\lambda\ \Longrightarrow\ 2\lambda=1\ \Longrightarrow\ \lambda=\frac12.

Step 3: Build that member. With $\lambda=\dfrac12$ :

x^2\!:\ 1+2=3,\qquad y^2\!:\ 2+1=3,\qquad x\!:\ -6-10=-16,\qquad y\!:\ -12-6=-18,

and the constant is $23+\dfrac{35}{2}=\dfrac{81}{2}$ .

Step 4: Normalise by dividing by $3$ .

x^2+y^2-\frac{16}{3}x-6y+\frac{27}{2}=0.

Step 5: Read the centre and radius. Comparing with $x^2+y^2+2gx+2fy+c=0$ gives $g=-\dfrac83$ , $f=-3$ , $c=\dfrac{27}{2}$ , so

(a,b)=\left(\frac83,\,3\right),\qquad r^2=g^2+f^2-c=\frac{64}{9}+9-\frac{27}{2}=\frac{128+162-243}{18}=\frac{47}{18}.

Step 6: Assemble.

ab+18r^2=\frac83\cdot3+18\cdot\frac{47}{18}=8+47=55.

Answer: $\boxed{55}$

Read this as a technique, not a special case

Whenever a question says "the intersection points of two curves lie on a circle", do not solve for the intersection points. Write $S_1+\lambda S_2=0$ and choose $\lambda$ to make it a circle. The four intersection points of two conics almost never have nice coordinates, but the circle through them nearly always does.

Type 3: circles through two given points

If the two points $A(x_1,y_1)$ and $B(x_2,y_2)$ are given directly, you do not need a circle to start from. The smallest circle through them is the one on $AB$ as diameter,

(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0,

and the line $AB$ can be written in determinant form as

\begin{vmatrix} x & y & 1\\ x_1 & y_1 & 1\\ x_2 & y_2 & 1\end{vmatrix}=0.

Adding $\lambda$ times the second to the first gives every circle through $A$ and $B$ :

(x-x_1)(x-x_2)+(y-y_1)(y-y_2)+\lambda\begin{vmatrix} x & y & 1\\ x_1 & y_1 & 1\\ x_2 & y_2 & 1\end{vmatrix}=0.

In practice you write the line $AB$ in ordinary form and add $\lambda$ times it. The determinant is just a way of remembering the line without a separate calculation.

Why the diameter circle works. If

P(x,y)

is any point of the circle on

AB

as diameter, other than

A

B

, then the angle in a semicircle gives

\angle APB=90^\circ

, so

PA\perp PB

. The vectors

\overrightarrow{PA}

and

\overrightarrow{PB}

have components

(x_1-x,\,y_1-y)

and

(x_2-x,\,y_2-y)

, and perpendicular vectors have zero dot product:

(x_1-x)(x_2-x)+(y_1-y)(y_2-y)=0.

Multiplying each bracket pair by $(-1)(-1)=1$ gives the form quoted above. It also holds at $P=A$ and $P=B$ , so the whole circle is captured.

Example 7. Find the circle through $A(1,-2)$ and $B(5,4)$ whose centre lies on the line $x-y=3$ .

▸ SHOW SOLUTION

Step 1: Write the diameter circle.

(x-1)(x-5)+(y+2)(y-4)=0\ \Longrightarrow\ x^2+y^2-6x-2y-3=0.

Step 2: Write the line $AB$ . Slope $=\dfrac{4+2}{5-1}=\dfrac32$ , so $y+2=\dfrac32(x-1)$ , that is

3x-2y-7=0.

Step 3: Write the family.

x^2+y^2-6x-2y-3+\lambda(3x-2y-7)=0.

Step 4: Find the centre. Comparing coefficients, $2g=-6+3\lambda$ and $2f=-2-2\lambda$ , so the centre is

\left(3-\frac{3\lambda}{2},\ 1+\lambda\right).

Step 5: Put the centre on $x-y=3$ .

\left(3-\frac{3\lambda}{2}\right)-(1+\lambda)=3\ \Longrightarrow\ 2-\frac{5\lambda}{2}=3\ \Longrightarrow\ \lambda=-\frac25.

Step 6: Substitute. Multiply the family by $5$ to clear fractions:

5(x^2+y^2-6x-2y-3)-2(3x-2y-7)=0\ \Longrightarrow\ 5x^2+5y^2-36x-6y-1=0.

Step 7: Check. The centre is $\left(\dfrac{18}{5},\dfrac35\right)$ and $\dfrac{18}{5}-\dfrac35=\dfrac{15}{5}=3$ , so it is on the given line. At $A(1,-2)$ the left side is $5+20-36+12-1=0$ , and at $B(5,4)$ it is $125+80-180-24-1=0$ .

Answer: $\boxed{5x^2+5y^2-36x-6y-1=0}$

Example 8. A circle passes through $A(2,0)$ and $B(0,4)$ and touches the $x$ -axis. Find it.

▸ SHOW SOLUTION

Step 1: The diameter circle.

(x-2)(x-0)+(y-0)(y-4)=0\ \Longrightarrow\ x^2+y^2-2x-4y=0.

Step 2: The line $AB$ . Intercept form $\dfrac x2+\dfrac y4=1$ gives $2x+y-4=0$ .

Step 3: The family.

x^2+y^2-2x-4y+\lambda(2x+y-4)=0,

that is $x^2+y^2+(2\lambda-2)x+(\lambda-4)y-4\lambda=0$ .

Step 4: Turn "touches the $x$ -axis" into an equation. Put $y=0$ and demand a repeated root:

x^2+(2\lambda-2)x-4\lambda=0,\qquad \text{discriminant}=0.

(2\lambda-2)^2+16\lambda=0\ \Longrightarrow\ 4\lambda^2-8\lambda+4+16\lambda=0\ \Longrightarrow\ 4\lambda^2+8\lambda+4=0.

Step 5: Solve. Dividing by $4$ , $(\lambda+1)^2=0$ , so $\lambda=-1$ .

Step 6: Substitute.

x^2+y^2-4x-5y+4=0.

Step 7: Check. Setting $y=0$ gives $x^2-4x+4=(x-2)^2=0$ , a repeated root at $x=2$ , so the circle touches the $x$ -axis at $(2,0)$ , which is $A$ itself. At $B(0,4)$ : $0+16-0-20+4=0$ . Correct.

Answer: $\boxed{x^2+y^2-4x-5y+4=0}$

Why only one answer here

A repeated root in $\lambda$ is the algebra telling you the two solutions have merged. Geometrically, $A$ already sits on the $x$ -axis, so any circle through $A$ that touches the $x$ -axis must touch it at $A$ , and there is only one such circle through $B$ as well. When $A$ is not on the axis you get a genuine quadratic in $\lambda$ and two circles.

Type 4: circles touching a given line at a given point

Suppose every circle must touch the line $L\equiv \ell x+my+n=0$ at the fixed point $T(x_1,y_1)$ , which lies on $L$ . Then

(x-x_1)^2+(y-y_1)^2+\lambda L=0.

This is the previous idea taken to its limit. The point circle $(x-x_1)^2+(y-y_1)^2=0$ plays the role of "the two intersection points have merged into one", and adding $\lambda L$ swings a whole family through that single point, all tangent to $L$ there.

Two special cases are worth keeping ready.

The line $L$ is	The family becomes
the vertical line $x=x_1$	$(x-x_1)^2+(y-y_1)^2+\lambda(x-x_1)=0$
the horizontal line $y=y_1$	$(x-x_1)^2+(y-y_1)^2+\lambda(y-y_1)=0$

Why it really is tangent. Substituting $L=0$ into the family leaves $(x-x_1)^2+(y-y_1)^2=0$ , whose only real solution is the single point $T$ . A line meeting a circle in exactly one point is a tangent, so every member touches $L$ at $T$ .

Example 9 (IIT JEE 2011). A circle passes through $(-1,0)$ and touches the $y$ -axis at $(0,2)$ . Which of $\left(-\dfrac32,0\right)$ , $\left(-\dfrac52,2\right)$ , $\left(-\dfrac32,\dfrac52\right)$ , $(-4,0)$ does it also pass through?

▸ SHOW SOLUTION

Step 1: Write the family. The tangency is at $(0,2)$ on the line $x=0$ , so

(x-0)^2+(y-2)^2+\lambda x=0.

Step 2: Impose the point $(-1,0)$ .

1+4-\lambda=0\ \Longrightarrow\ \lambda=5.

Step 3: Write the circle.

x^2+y^2+5x-4y+4=0.

Step 4: Test the four options.

$\left(-\dfrac32,0\right)$ : $\dfrac94-\dfrac{15}{2}+4=\dfrac{9-30+16}{4}=-\dfrac54\neq0$ .
$\left(-\dfrac52,2\right)$ : $\dfrac{25}{4}+4-\dfrac{25}{2}-8+4=\dfrac{25+16-50-32+16}{4}=-\dfrac{25}{4}\neq0$ .
$\left(-\dfrac32,\dfrac52\right)$ : $\dfrac94+\dfrac{25}{4}-\dfrac{15}{2}-10+4=\dfrac{9+25-30-40+16}{4}=-5\neq0$ .
$(-4,0)$ : $16+0-20-0+4=0$ . This one works.

Step 5: Check the tangency. Setting $x=0$ gives $y^2-4y+4=(y-2)^2=0$ , a repeated root, so the circle really does touch the $y$ -axis at $(0,2)$ .

Answer: $\boxed{(-4,0)}$

Example 10. Find the radius of the smallest circle that touches the line $3x-y-6=0$ at $(1,-3)$ and also touches the line $y=x$ .

▸ SHOW SOLUTION

Step 1: Confirm the point is on the line. $3(1)-(-3)-6=3+3-6=0$ . Yes.

Step 2: Write the family.

(x-1)^2+(y+3)^2+\lambda(3x-y-6)=0,

which expands to

x^2+y^2+(3\lambda-2)x+(6-\lambda)y+(10-6\lambda)=0.

Step 3: Impose tangency to $y=x$ . Substituting $y=x$ :

2x^2+(3\lambda-2+6-\lambda)x+(10-6\lambda)=0\ \Longrightarrow\ 2x^2+(2\lambda+4)x+(10-6\lambda)=0.

Dividing by $2$ :

x^2+(\lambda+2)x+(5-3\lambda)=0.

Step 4: Set the discriminant to zero.

(\lambda+2)^2-4(5-3\lambda)=0\ \Longrightarrow\ \lambda^2+4\lambda+4-20+12\lambda=0\ \Longrightarrow\ \lambda^2+16\lambda-16=0.

Step 5: Solve. $\lambda=\dfrac{-16\pm\sqrt{256+64}}{2}=-8\pm\sqrt{80}=-8\pm4\sqrt5$ .

Step 6: Find the radius. For the general member, $g=\dfrac{3\lambda-2}{2}$ , $f=\dfrac{6-\lambda}{2}$ , $c=10-6\lambda$ , so

r^2=g^2+f^2-c=\frac{(3\lambda-2)^2+(6-\lambda)^2-4(10-6\lambda)}{4}=\frac{9\lambda^2-12\lambda+4+36-12\lambda+\lambda^2-40+24\lambda}{4}=\frac{10\lambda^2}{4}.

So $r=\dfrac{|\lambda|\sqrt{10}}{2}$ , and the smaller circle comes from the $\lambda$ of smaller absolute value, namely $\lambda=-8+4\sqrt5\approx0.944$ .

Step 7: Compute.

r=\frac{(4\sqrt5-8)\sqrt{10}}{2}=(2\sqrt5-4)\sqrt{10}=10\sqrt2-4\sqrt{10}\approx14.14-12.65\approx1.5.

Answer: $\boxed{r\approx1.5}$

Radius straight from the family

Notice that Step 6 gave $r^2$ as a clean function of $\lambda$ before either root was substituted. Doing the general computation once, then plugging in, is nearly always shorter than building both circles and measuring them separately.

Which form to reach for

All four types are the same move. Write down something that already satisfies the conditions you were given, attach one $\lambda$ , and let the last condition fix it.

The circle must pass through or touch	Write
the intersection of a circle $S=0$ and a line $L=0$	$S+\lambda L=0$
the intersection of two circles $S_1=0$ , $S_2=0$	$S_1+\lambda S_2=0$ , with $\lambda\neq-1$
two given points $A(x_1,y_1)$ , $B(x_2,y_2)$	$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)+\lambda\,(\text{line }AB)=0$
a line $L=0$ at a given point $T(x_1,y_1)$ on it	$(x-x_1)^2+(y-y_1)^2+\lambda L=0$

And then the last condition, whatever it is, becomes one equation in $\lambda$ :

Last condition	Equation in $\lambda$
passes through a further point	substitute the point
centre on a given line	put the centre expression into that line
$AB$ is a diameter, or the circle is the smallest of the family	centre lies on the chord
touches a given line	discriminant of the substituted quadratic is zero
given radius	$g^2+f^2-c=r^2$
cuts a given circle orthogonally	$2(g_1g_2+f_1f_2)=c_1+c_2$
bisects the circumference of a given circle	common chord passes through that circle's centre

One habit that saves time in the exam

Before substituting, always write the family in the normalised form $x^2+y^2+2gx+2fy+c=0$ and note $g$ , $f$ , $c$ as expressions in $\lambda$ . Every condition in the second table above is then a one-line substitution, and you never rebuild the circle twice.

The common chord

We met it as the degenerate member of the family, and it deserves its own name. For two intersecting circles

S_1\equiv x^2+y^2+2g_1x+2f_1y+c_1=0,\qquad S_2\equiv x^2+y^2+2g_2x+2f_2y+c_2=0,

the chord joining their common points is

S_1-S_2=0,\qquad\text{that is}\qquad 2(g_1-g_2)x+2(f_1-f_2)y+(c_1-c_2)=0.

Length. Drop a perpendicular from $C_1$ to the chord. It bisects the chord, so with $d_1$ the distance from $C_1$ to the line $S_1-S_2=0$ ,

PQ=2\sqrt{r_1^2-d_1^2}.

You may use either centre. The answer is the same, which is a good check.

The line exists even when the circles do not meet

$S_1-S_2=0$ is a perfectly good line whether or not the circles intersect. It is called the radical axis, and it is always perpendicular to the line of centres. When the circles meet it is the common chord, when they touch it is the common tangent at the point of contact, and when they are apart it is a line lying between them. Every member of the family $S_1+\lambda S_2=0$ shares that same radical axis with both, which is why the family is called coaxial.

Example 11. Find the length of the common chord of $x^2+y^2+8x+1=0$ and $x^2+y^2+2\mu y-1=0$ , given that it equals $2\sqrt6$ , and hence find $\mu$ .

▸ SHOW SOLUTION

Step 1: Write the common chord. Subtracting,

8x-2\mu y+2=0\ \Longrightarrow\ 4x-\mu y+1=0.

Step 2: Take the first circle. Here $g=4$ , $f=0$ , $c=1$ , so its centre is $(-4,0)$ and

r_1=\sqrt{g^2+f^2-c}=\sqrt{16+0-1}=\sqrt{15}.

Step 3: Use the half-chord relation. Half the chord is $\sqrt6$ , so

d_1^2=r_1^2-6=15-6=9\ \Longrightarrow\ d_1=3.

Step 4: Compute $d_1$ from the line. The distance from $(-4,0)$ to $4x-\mu y+1=0$ is

d_1=\frac{|4(-4)-\mu(0)+1|}{\sqrt{16+\mu^2}}=\frac{15}{\sqrt{16+\mu^2}}.

Step 5: Solve.

\frac{15}{\sqrt{16+\mu^2}}=3\ \Longrightarrow\ \sqrt{16+\mu^2}=5\ \Longrightarrow\ \mu^2=9\ \Longrightarrow\ \mu=\pm3.

Step 6: Check with the other circle. For $\mu=3$ the second circle has centre $(0,-3)$ and radius $\sqrt{0+9+1}=\sqrt{10}$ . Its distance to $4x-3y+1=0$ is $\dfrac{|0+9+1|}{5}=2$ , and $2\sqrt{10-4}=2\sqrt6$ . It matches.

Answer: $\boxed{\mu=\pm3}$

Example 12. If the circle $x^2+y^2+4x+22y+c=0$ bisects the circumference of $x^2+y^2-2x+8y-d=0$ , find $c+d$ .

▸ SHOW SOLUTION

Step 1: Turn "bisects the circumference" into geometry. One circle bisects the circumference of another exactly when their common chord is a diameter of the second circle, so the common chord passes through the centre of the second circle.

Step 2: Write the common chord. Subtracting the second from the first,

(4+2)x+(22-8)y+(c+d)=0\ \Longrightarrow\ 6x+14y+c+d=0.

Step 3: Locate the second centre. For $x^2+y^2-2x+8y-d=0$ , the centre is $(1,-4)$ .

Step 4: Put the centre on the chord.

6(1)+14(-4)+c+d=0\ \Longrightarrow\ 6-56+c+d=0.

Step 5: Solve.

c+d=50.

Answer: $\boxed{50}$

Bisecting is not symmetric

" $S_1$ bisects the circumference of $S_2$ " is a statement about $S_2$ 's centre, not $S_1$ 's. Substituting the wrong centre is the standard way to lose this mark. Read the sentence, then ask: whose circumference is being cut in half? That circle's centre is the one that goes into the chord.

Example 13 (JEE Main 2026, 23 Jan Shift 2). Find the area of the region common to the circles $x^2+y^2=4$ and $x^2+(y-2)^2=4$ .

▸ SHOW SOLUTION

Step 1: Find the common chord. Subtracting,

y^2-(y-2)^2=0\ \Longrightarrow\ 4y-4=0\ \Longrightarrow\ y=1.

Step 2: Find where it meets the circles. Putting $y=1$ into $x^2+y^2=4$ gives $x^2=3$ , so the circles meet at $(\pm\sqrt3,1)$ .

Step 3: Set up the area. Over $-\sqrt3\le x\le\sqrt3$ the region lies below the upper circle $y=\sqrt{4-x^2}$ and above the lower boundary $y=2-\sqrt{4-x^2}$ . Using symmetry in $x$ ,

A=2\int_0^{\sqrt3}\Big[\sqrt{4-x^2}-\big(2-\sqrt{4-x^2}\big)\Big]dx=4\int_0^{\sqrt3}\Big(\sqrt{4-x^2}-1\Big)dx.

Step 4: Integrate. Using $\displaystyle\int\sqrt{a^2-x^2}\,dx=\frac x2\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac xa$ with $a=2$ ,

A=4\left[\frac x2\sqrt{4-x^2}+2\sin^{-1}\frac x2-x\right]_0^{\sqrt3}.

Step 5: Evaluate. At $x=\sqrt3$ : $\sqrt{4-3}=1$ and $\sin^{-1}\dfrac{\sqrt3}{2}=\dfrac\pi3$ , so the bracket is $\dfrac{\sqrt3}{2}+\dfrac{2\pi}{3}-\sqrt3=\dfrac{2\pi}{3}-\dfrac{\sqrt3}{2}$ .

Step 6: Multiply out.

A=4\left(\frac{2\pi}{3}-\frac{\sqrt3}{2}\right)=\frac{8\pi}{3}-2\sqrt3=\frac23\left(4\pi-3\sqrt3\right).

Step 7: Sanity check. Numerically $A\approx8.378-3.464\approx4.91$ , which is a sensible fraction of a single circle's area $4\pi\approx12.57$ .

Answer: $\boxed{\dfrac23\left(4\pi-3\sqrt3\right)}$

Two circles in contact

Two circles touch when they meet at exactly one point. Everything follows from one comparison: the distance between the centres against the two radii.

\text{external contact: } C_1C_2=r_1+r_2,\qquad \text{internal contact: } C_1C_2=|r_1-r_2|.

The point of contact $P$ always lies on the line $C_1C_2$ , and the section formula gives it at once.

Contact	$P$ divides $C_1C_2$	Coordinates of $P$
external	internally in $r_1:r_2$	$\left(\dfrac{r_1x_2+r_2x_1}{r_1+r_2},\ \dfrac{r_1y_2+r_2y_1}{r_1+r_2}\right)$
internal	externally in $r_1:r_2$	$\left(\dfrac{r_1x_2-r_2x_1}{r_1-r_2},\ \dfrac{r_1y_2-r_2y_1}{r_1-r_2}\right)$

Example 14 (IIT JEE 2014). Let $C$ be the circle with centre $(1,1)$ and radius $1$ . If $T$ is the circle centred at $(0,y)$ passing through the origin and touching $C$ externally, find the radius of $T$ .

▸ SHOW SOLUTION

Step 1: Name the radius. $T$ has centre $(0,k)$ and passes through the origin, so its radius is $|k|$ . Take $k>0$ .

Step 2: Write the contact condition. External contact means the distance between centres equals the sum of radii:

\sqrt{(1-0)^2+(1-k)^2}=k+1.

Step 3: Square both sides.

1+(1-k)^2=(k+1)^2\ \Longrightarrow\ 1+1-2k+k^2=k^2+2k+1.

Step 4: Simplify.

2-2k=2k+1\ \Longrightarrow\ 1=4k\ \Longrightarrow\ k=\frac14.

Step 5: Check. Centres $(1,1)$ and $\left(0,\dfrac14\right)$ are $\sqrt{1+\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac54$ apart, and $r_1+r_2=1+\dfrac14=\dfrac54$ . It matches.

Answer: $\boxed{\dfrac14}$

Example 15 (JEE Main 2026, 2 Apr Shift 2). Let $A=(3,0)$ . Circles with variable diameter $AB$ touch the circle $x^2+y^2=36$ internally. If the locus of $B$ is a curve of eccentricity $e$ , find $72e^2$ .

▸ SHOW SOLUTION

Step 1: Set up. Write $B=(h,k)$ . The variable circle has $AB$ as a diameter, so its centre is the midpoint

M=\left(\frac{h+3}{2},\ \frac{k}{2}\right),

and its radius is

\rho=\frac{AB}{2}=\frac{\sqrt{(h-3)^2+k^2}}{2}.

Step 2: Write the internal contact condition. The fixed circle has centre $O(0,0)$ and radius $6$ . The small circle sits inside it, so

OM=6-\rho.

Step 3: Substitute.

\sqrt{\left(\frac{h+3}{2}\right)^2+\left(\frac{k}{2}\right)^2}=6-\frac{\sqrt{(h-3)^2+k^2}}{2}.

Step 4: Multiply through by $2$ .

\sqrt{(h+3)^2+k^2}=12-\sqrt{(h-3)^2+k^2}.

Step 5: Rearrange and read the definition.

\sqrt{(h+3)^2+k^2}+\sqrt{(h-3)^2+k^2}=12.

The sum of the distances from $B$ to the two fixed points $(-3,0)$ and $(3,0)$ is constant. That is precisely an ellipse with those two points as foci.

Step 6: Extract the constants. $2a=12$ gives $a=6$ , and the foci are $6$ apart so $2c=6$ , giving $c=3$ .

Step 7: Compute the eccentricity.

e=\frac ca=\frac36=\frac12,\qquad 72e^2=72\cdot\frac14=18.

Answer: $\boxed{18}$

Recognising a conic from a contact condition

Contact conditions turn into sums or differences of two distances, which is exactly how ellipses and hyperbolas are defined. Internal contact with a fixed circle gives a sum of distances, so an ellipse. External contact with two fixed circles gives a difference, so a hyperbola. Spot the sum or difference and you can name the curve before doing any algebra.

Common tangents: counting them first

A common tangent touches both circles. How many exist depends only on how the circles sit relative to each other, and the whole classification is one comparison of $C_1C_2$ with $r_1+r_2$ and $|r_1-r_2|$ .

Position	Condition	Common tangents
one outside the other	$C_1C_2>r_1+r_2$	$4$
touching externally	$C_1C_2=r_1+r_2$	$3$
cutting at two points	$	r_1-r_2
touching internally	$C_1C_2=	r_1-r_2
one inside the other	$C_1C_2<	r_1-r_2

The two tangents that keep both circles on the same side are the direct common tangents, and they meet on the line of centres at the point dividing $C_1C_2$ externally in the ratio $r_1:r_2$ . The two that pass between the circles are the transverse common tangents, and they meet at the point dividing $C_1C_2$ internally in the same ratio.

With $d=C_1C_2$ , the tangent segments have lengths

\text{direct}=\sqrt{d^2-(r_1-r_2)^2},\qquad \text{transverse}=\sqrt{d^2-(r_1+r_2)^2}.

The second formula makes the classification obvious: the transverse tangents are real only when $d\ge r_1+r_2$ , which is exactly the case where you counted $3$ or $4$ .

Example 16 (JEE Main 2015). How many common tangents can be drawn to $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ ?

▸ SHOW SOLUTION

Step 1: Read off centres and radii. For the first, $g=-2$ , $f=-3$ , $c=-12$ , so $C_1=(2,3)$ and

r_1=\sqrt{4+9+12}=\sqrt{25}=5.

For the second, $g=3$ , $f=9$ , $c=26$ , so $C_2=(-3,-9)$ and

r_2=\sqrt{9+81-26}=\sqrt{64}=8.

Step 2: Compute the distance between centres.

C_1C_2=\sqrt{(2+3)^2+(3+9)^2}=\sqrt{25+144}=\sqrt{169}=13.

Step 3: Compare. $r_1+r_2=5+8=13=C_1C_2$ , so the circles touch externally.

Step 4: Read the table. External contact gives $3$ common tangents.

Answer: $\boxed{3}$

Example 17 (JEE Main 2026, 22 Jan Shift 1). The set of values of $r$ for which the circles $(x+1)^2+(y+4)^2=r^2$ and $x^2+y^2-4x-2y-4=0$ intersect at two distinct points is an interval $(\alpha,\beta)$ . Find $\alpha\beta$ .

▸ SHOW SOLUTION

Step 1: Put the second circle in standard form. Completing squares,

(x-2)^2+(y-1)^2=4+4+1=9,

so $C_2=(2,1)$ and $r_2=3$ . The first has $C_1=(-1,-4)$ and radius $r$ .

Step 2: Compute the distance between centres.

d=\sqrt{(2+1)^2+(1+4)^2}=\sqrt{9+25}=\sqrt{34}.

Step 3: Write the intersecting condition. Two circles cut at two distinct points exactly when

|r-3|<d<r+3.

Step 4: Solve the right half. $\sqrt{34}<r+3$ gives $r>\sqrt{34}-3$ .

Step 5: Solve the left half. $|r-3|<\sqrt{34}$ gives $-\sqrt{34}<r-3<\sqrt{34}$ , and since $r>0$ the binding part is $r<\sqrt{34}+3$ .

Step 6: Combine. $r\in\left(\sqrt{34}-3,\ \sqrt{34}+3\right)$ , so $\alpha=\sqrt{34}-3$ and $\beta=\sqrt{34}+3$ .

Step 7: Multiply.

\alpha\beta=\left(\sqrt{34}\right)^2-3^2=34-9=25.

Answer: $\boxed{25}$

Example 18. Find the length of the shortest segment $PQ$ that touches $C_1: x^2+y^2-20x+64=0$ at $P$ and $C_2: x^2+y^2+30x+144=0$ at $Q$ .

▸ SHOW SOLUTION

Step 1: Read off the two circles. For $C_1$ : centre $(10,0)$ and $r_1=\sqrt{100-64}=6$ . For $C_2$ : centre $(-15,0)$ and $r_2=\sqrt{225-144}=9$ .

Step 2: Check their position. $d=25$ and $r_1+r_2=15<25$ , so the circles lie completely outside each other and all four common tangents are real.

Step 3: Identify what "shortest" means. A segment touching both circles is a piece of a common tangent. The direct tangent segments have length $\sqrt{d^2-(r_1-r_2)^2}$ and the transverse ones $\sqrt{d^2-(r_1+r_2)^2}$ . Since $(r_1+r_2)^2>(r_1-r_2)^2$ , the transverse segment is the shorter one.

Step 4: Compute.

PQ=\sqrt{25^2-(6+9)^2}=\sqrt{625-225}=\sqrt{400}=20.

Step 5: Check the other one. The direct segment would be $\sqrt{625-9}=\sqrt{616}\approx24.8$ , indeed longer.

Answer: $\boxed{20}$

Do not memorise which formula is which

Students routinely swap the two tangent lengths. Rebuild them instead: the transverse tangents pass between the circles, so they must clear the sum of the radii, hence $r_1+r_2$ appears. The direct tangents run alongside, so only the mismatch in radii matters, hence $r_1-r_2$ . Reasoning takes two seconds and never fails.

Angle of intersection and orthogonal circles

When two circles cut, the angle between them is defined as the angle between their tangents at a common point. Since each tangent is perpendicular to the corresponding radius, that is the same as the angle between $C_1P$ and $C_2P$ .

In triangle $C_1PC_2$ the sides are $r_1$ , $r_2$ and $d=C_1C_2$ , so the cosine rule gives

\cos\theta=\frac{r_1^2+r_2^2-d^2}{2r_1r_2}.

Orthogonal circles. Setting $\theta=90^\circ$ makes the numerator vanish:

r_1^2+r_2^2=d^2.

Written in general coefficients, with $r_1^2=g_1^2+f_1^2-c_1$ , $r_2^2=g_2^2+f_2^2-c_2$ and $d^2=(g_1-g_2)^2+(f_1-f_2)^2$ , everything squared cancels and what survives is

\boxed{\,2(g_1g_2+f_1f_2)=c_1+c_2\,}

A picture worth carrying

Orthogonality says $C_1P\perp C_2P$ , and since $C_2P$ is a radius of the second circle while $C_1P$ is perpendicular to it, $C_1P$ is tangent to the second circle at $P$ . So two circles are orthogonal exactly when each radius drawn to a common point is a tangent to the other circle. That reading turns many orthogonality questions into tangent-length questions.

Example 19. Find the angle at which $x^2+y^2-6x+4y+11=0$ and $x^2+y^2-4x+6y+9=0$ intersect.

▸ SHOW SOLUTION

Step 1: Read off centres and radii. First: $C_1=(3,-2)$ , $r_1=\sqrt{9+4-11}=\sqrt2$ . Second: $C_2=(2,-3)$ , $r_2=\sqrt{4+9-9}=2$ .

Step 2: Compute the distance between centres.

d=\sqrt{(3-2)^2+(-2+3)^2}=\sqrt{1+1}=\sqrt2.

Step 3: Apply the cosine rule.

\cos\theta=\frac{r_1^2+r_2^2-d^2}{2r_1r_2}=\frac{2+4-2}{2\cdot\sqrt2\cdot2}=\frac{4}{4\sqrt2}=\frac{1}{\sqrt2}.

Step 4: Read the angle. $\theta=45^\circ$ .

Step 5: Check they really do meet. $|r_1-r_2|=2-\sqrt2\approx0.586$ and $r_1+r_2=2+\sqrt2\approx3.414$ , and $d=\sqrt2\approx1.414$ lies strictly between them, so the circles genuinely cut at two points.

Answer: $\boxed{45^\circ}$

Example 20. For what value of $k$ do $x^2+y^2-2kx+6y+1=0$ and $x^2+y^2+4x-2y+3=0$ cut orthogonally?

▸ SHOW SOLUTION

Step 1: Read off the coefficients. First: $g_1=-k$ , $f_1=3$ , $c_1=1$ . Second: $g_2=2$ , $f_2=-1$ , $c_2=3$ .

Step 2: Apply the orthogonality condition.

2(g_1g_2+f_1f_2)=c_1+c_2\ \Longrightarrow\ 2\big((-k)(2)+(3)(-1)\big)=1+3.

Step 3: Simplify.

2(-2k-3)=4\ \Longrightarrow\ -4k-6=4\ \Longrightarrow\ k=-\frac52.

Step 4: Check both circles are real. With $k=-\dfrac52$ , the first has $r_1^2=\dfrac{25}{4}+9-1=\dfrac{57}{4}>0$ and the second has $r_2^2=4+1-3=2>0$ . Both are genuine circles.

Step 5: Verify geometrically. The centres are $\left(-\dfrac52,-3\right)$ and $(-2,1)$ , so

d^2=\left(\frac12\right)^2+4^2=\frac14+16=\frac{65}{4},\qquad r_1^2+r_2^2=\frac{57}{4}+2=\frac{65}{4}.

They are equal, which is exactly the orthogonality condition $r_1^2+r_2^2=d^2$ .

Answer: $\boxed{k=-\dfrac52}$

Example 21 (JEE Advanced 2014). A circle $S$ passes through $(0,1)$ and is orthogonal to $(x-1)^2+y^2=16$ and $x^2+y^2=1$ . Find its centre and radius.

▸ SHOW SOLUTION

Step 1: Name the unknown circle. Let $S: x^2+y^2+2gx+2fy+c=0$ .

Step 2: Write the two given circles in general form. $(x-1)^2+y^2=16$ becomes $x^2+y^2-2x-15=0$ , so $g_2=-1$ , $f_2=0$ , $c_2=-15$ . And $x^2+y^2=1$ becomes $x^2+y^2-1=0$ , so $g_3=0$ , $f_3=0$ , $c_3=-1$ .

Step 3: Orthogonality with the second circle.

2(g\cdot0+f\cdot0)=c+(-1)\ \Longrightarrow\ 0=c-1\ \Longrightarrow\ c=1.

Step 4: Orthogonality with the first circle.

2(g\cdot(-1)+f\cdot0)=c+(-15)\ \Longrightarrow\ -2g=1-15=-14\ \Longrightarrow\ g=7.

Step 5: Use the point $(0,1)$ .

0+1+0+2f+c=0\ \Longrightarrow\ 1+2f+1=0\ \Longrightarrow\ f=-1.

Step 6: Assemble. The centre is $(-g,-f)=(-7,1)$ and

r=\sqrt{g^2+f^2-c}=\sqrt{49+1-1}=\sqrt{49}=7.

Step 7: Check. Distance from $(-7,1)$ to the centre $(1,0)$ of the first given circle is $\sqrt{64+1}=\sqrt{65}$ , and $r^2+4^2=49+16=65$ . Orthogonal. Distance to $(0,0)$ is $\sqrt{50}$ , and $49+1=50$ . Orthogonal again.

Answer: $\boxed{\text{centre }(-7,1),\ \text{radius }7}$

Chords and loci built on the same ideas

The three questions below are all from JEE Main 2026. None of them says "family of circles", and all three are solved by the habits this chapter builds: name the moving point, write the geometric condition, and let the algebra become a circle.

Example 22 (JEE Main 2026, 8 Apr Shift 2). For the circle $C: x^2+y^2-6x-8y-11=0$ , a variable chord $AB$ subtends a right angle at the origin. The locus of the foot of the perpendicular from the origin to $AB$ is the circle $x^2+y^2-\alpha x-\beta y-\gamma=0$ . Find $\alpha+\beta+2\gamma$ .

▸ SHOW SOLUTION

Step 1: Read the given circle. Its centre is $C=(3,4)$ and

R=\sqrt{9+16+11}=\sqrt{36}=6.

Step 2: Name the moving point and the midpoint. Let $P(h,k)$ be the foot of the perpendicular from $O$ to $AB$ , and let $M$ be the midpoint of $AB$ . Since $CM\perp AB$ , the point $M$ is the foot of the perpendicular from $C$ .

Step 3: Use the right angle at $O$ . In triangle $AOB$ the angle at $O$ is $90^\circ$ , so $AB$ is the hypotenuse and the midpoint of the hypotenuse is equidistant from all three vertices:

OM=AM=\frac{AB}{2}.

Step 4: Express $AM$ from the circle. In the right triangle $AMC$ ,

AM^2=R^2-CM^2.

Step 5: Express $OM$ using $P$ . Both $M$ and $P$ lie on the line $AB$ , and $OP\perp AB$ , so triangle $OPM$ is right angled at $P$ :

OM^2=OP^2+PM^2.

Similarly triangle $CPM$ is right angled at $M$ , giving $CP^2=CM^2+MP^2$ , that is $CM^2=CP^2-MP^2$ .

Step 6: Put the three together. Substituting Steps 4 and 5 into $OM^2=AM^2$ :

OP^2+PM^2=R^2-\left(CP^2-MP^2\right).

The $PM^2$ terms cancel and what is left is beautifully simple:

OP^2+CP^2=R^2.

Step 7: Write that in coordinates. With $P=(h,k)$ , $O=(0,0)$ , $C=(3,4)$ and $R=6$ :

h^2+k^2+(h-3)^2+(k-4)^2=36.

Step 8: Expand and tidy.

2h^2+2k^2-6h-8k+25=36\ \Longrightarrow\ 2h^2+2k^2-6h-8k-11=0.

Dividing by $2$ ,

h^2+k^2-3h-4k-\frac{11}{2}=0.

Step 9: Compare with the given form. So $\alpha=3$ , $\beta=4$ and $\gamma=\dfrac{11}{2}$ , giving

\alpha+\beta+2\gamma=3+4+11=18.

Answer: $\boxed{18}$

The one fact that did all the work

"The midpoint of the hypotenuse of a right triangle is equidistant from all three vertices" is a class 9 circle theorem, and it converted a right-angle condition into a length equation in a single line. Whenever a chord subtends $90^\circ$ at a point, reach for it first.

Example 23 (JEE Main 2026, 4 Apr Shift 1). Two chords are drawn from the point $(1,2)$ on the circle $x^2+y^2+x-3y=0$ and each is bisected by the $y$ -axis. If the far ends of these chords are $R$ and $S$ , and the midpoint of $RS$ is $(\alpha,\beta)$ , find $6(\alpha+\beta)$ .

▸ SHOW SOLUTION

Step 1: Check the given point is on the circle. $1+4+1-6=0$ . Yes.

Step 2: Parametrise the midpoints. A chord bisected by the $y$ -axis has its midpoint at $M(0,\lambda)$ for some $\lambda$ .

Step 3: Find the far end. If $P(1,2)$ and $E$ are the ends of a chord with midpoint $M$ , then $M=\dfrac{P+E}{2}$ , so

E=2M-P=(-1,\ 2\lambda-2).

Step 4: Force $E$ onto the circle. Substituting $x=-1$ , $y=2\lambda-2$ into $x^2+y^2+x-3y=0$ :

1+(2\lambda-2)^2-1-3(2\lambda-2)=0.

Step 5: Expand.

4\lambda^2-8\lambda+4-6\lambda+6=0\ \Longrightarrow\ 4\lambda^2-14\lambda+10=0.

Step 6: Solve. Dividing by $2$ , $2\lambda^2-7\lambda+5=0$ , so $(2\lambda-5)(\lambda-1)=0$ and $\lambda=1$ or $\lambda=\dfrac52$ .

Step 7: Get the two far ends. For $\lambda=1$ : $E=(-1,0)$ . For $\lambda=\dfrac52$ : $E=(-1,3)$ . So $R=(-1,0)$ and $S=(-1,3)$ .

Step 8: Take the midpoint of $RS$ .

(\alpha,\beta)=\left(-1,\ \frac32\right),\qquad 6(\alpha+\beta)=6\left(-1+\frac32\right)=6\cdot\frac12=3.

Step 9: Check. Both $(-1,0)$ and $(-1,3)$ satisfy $x^2+y^2+x-3y=0$ : $1+0-1-0=0$ and $1+9-1-9=0$ .

Answer: $\boxed{3}$

Example 24 (JEE Main 2026, 6 Apr Shift 2). The line $x-y=4$ meets the circle $C:(x-4)^2+(y+3)^2=9$ at $Q$ and $R$ . If $P(\alpha,\beta)$ lies on $C$ with $PQ=PR$ , find $(6\alpha+8\beta)^2$ .

▸ SHOW SOLUTION

Step 1: Find $Q$ and $R$ . Substituting $y=x-4$ :

(x-4)^2+(x-1)^2=9\ \Longrightarrow\ 2x^2-10x+8=0\ \Longrightarrow\ x^2-5x+4=0,

so $x=1$ or $x=4$ , giving $Q=(4,0)$ and $R=(1,-3)$ .

Step 2: Turn $PQ=PR$ into a line. Points equidistant from $Q$ and $R$ lie on the perpendicular bisector of $QR$ . Writing it out,

(\alpha-4)^2+\beta^2=(\alpha-1)^2+(\beta+3)^2.

Step 3: Expand and cancel.

-8\alpha+16=-2\alpha+1+6\beta+9\ \Longrightarrow\ -6\alpha-6\beta+6=0\ \Longrightarrow\ \alpha+\beta=1.

Step 4: Use the relation directly. We need $6\alpha+8\beta$ , and $\beta=1-\alpha$ , so

6\alpha+8\beta=6\alpha+8-8\alpha=8-2\alpha.

Step 5: Find $\alpha$ . $P$ is on the circle, so substituting $\beta=1-\alpha$ into $(\alpha-4)^2+(\beta+3)^2=9$ :

(\alpha-4)^2+(4-\alpha)^2=9\ \Longrightarrow\ 2(\alpha-4)^2=9\ \Longrightarrow\ \alpha=4\pm\frac{3}{\sqrt2}.

Step 6: Substitute.

6\alpha+8\beta=8-2\left(4\pm\frac{3}{\sqrt2}\right)=\mp\frac{6}{\sqrt2},

so in either case

(6\alpha+8\beta)^2=\frac{36}{2}=18.

Step 7: Check. The perpendicular bisector of $QR$ passes through the centre $(4,-3)$ , and indeed $4+(-3)=1$ . Both solutions are the two points where that bisector meets the circle, and the squaring makes the sign irrelevant.

Answer: $\boxed{18}$

Quick recall

Item	Result
Family of lines through the intersection of $L_1,L_2$	$L_1+\lambda L_2=0$
Fixed point of a family $A+\lambda B=0$	solve $A=0$ and $B=0$
Circles through a circle and a line	$S+\lambda L=0$
Circles through two circles	$S_1+\lambda S_2=0$ , $\lambda\neq-1$
Circle on $AB$ as diameter	$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$
Circles through two points $A,B$	diameter circle $+\ \lambda\,(\text{line }AB)=0$
Circles touching $L$ at $T(x_1,y_1)$	$(x-x_1)^2+(y-y_1)^2+\lambda L=0$
Common chord / radical axis	$S_1-S_2=0$
Length of the common chord	$2\sqrt{r_1^2-d_1^2}$ , $d_1$ from $C_1$ to the chord
$S_1$ bisects the circumference of $S_2$	common chord passes through the centre of $S_2$
External contact	$C_1C_2=r_1+r_2$ , $P$ divides $C_1C_2$ internally in $r_1:r_2$
Internal contact	$C_1C_2=\lvert r_1-r_2\rvert$ , $P$ divides externally in $r_1:r_2$
Number of common tangents	$4$ , $3$ , $2$ , $1$ , $0$ as $C_1C_2$ falls through $r_1+r_2$ and $\lvert r_1-r_2\rvert$
Direct common tangent length	$\sqrt{d^2-(r_1-r_2)^2}$
Transverse common tangent length	$\sqrt{d^2-(r_1+r_2)^2}$
Angle of intersection	$\cos\theta=\dfrac{r_1^2+r_2^2-d^2}{2r_1r_2}$
Orthogonality	$2(g_1g_2+f_1f_2)=c_1+c_2$ , equivalently $r_1^2+r_2^2=d^2$
Chord subtending $90^\circ$ at $O$ , foot $P$	$OP^2+CP^2=R^2$

The one habit to carry into the exam

Count the conditions before you compute anything. Three conditions means a unique circle and you should be solving for $g$ , $f$ , $c$ . Two conditions means a family, so write the family in one line and keep $\lambda$ alive until the last condition arrives. Deciding this in the first ten seconds is worth more than any formula on this page.

The five slips that cost marks in this chapter

Not normalising the coefficient of $x^2$ to $1$ before combining. Forgetting $\lambda\neq-1$ in $S_1+\lambda S_2=0$ . Swapping the direct and transverse tangent lengths. Substituting the wrong centre in a "bisects the circumference" question. Confusing the longest chord through a point with the chord farthest from the centre.

Work every one of these on paper before you look at the solutions. Then come back to the doMath practice sets on Circles and time yourself: two minutes for a tangent-counting question, four for a family question, five for a locus.

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Reinforce this post with live problems on doMath.

Straight Lines Circles Parabola Ellipse Hyperbola

Family of Circles: Complete JEE Notes

Family of lines: where the λ\lambdaλ trick starts

Type 1: circles through the intersection of a circle and a line

Step 1: Write the family.

Step 4: Substitute back.

Step 4: Put the centre on y=xy=xy=x.

Step 5: Substitute.

Type 2: circles through the intersection of two circles

Step 1: Write the family.

Step 4: Substitute and divide by 1+λ=21+\lambda=21+λ=2.

Step 4: Normalise by dividing by 333.

Step 6: Assemble.

Type 3: circles through two given points

Step 1: Write the diameter circle.

Step 3: Write the family.

Step 5: Put the centre on x−y=3x-y=3x−y=3.

Step 1: The diameter circle.

Step 3: The family.

Step 6: Substitute.

Type 4: circles touching a given line at a given point

Step 2: Impose the point (−1,0)(-1,0)(−1,0).

Step 3: Write the circle.

Step 4: Test the four options.

Step 2: Write the family.

Step 4: Set the discriminant to zero.

Step 7: Compute.

Which form to reach for

The common chord

Step 5: Solve.

Step 4: Put the centre on the chord.

Step 5: Solve.

Step 6: Multiply out.

Two circles in contact

Step 3: Square both sides.

Step 4: Simplify.

Step 3: Substitute.

Step 4: Multiply through by 222.

Step 5: Rearrange and read the definition.

Step 7: Compute the eccentricity.

Common tangents: counting them first

Step 2: Compute the distance between centres.

Step 2: Compute the distance between centres.

Step 7: Multiply.

Step 4: Compute.

Angle of intersection and orthogonal circles

Step 2: Compute the distance between centres.

Step 3: Apply the cosine rule.

Step 2: Apply the orthogonality condition.

Step 3: Simplify.

Step 3: Orthogonality with the second circle.

Step 4: Orthogonality with the first circle.

Step 5: Use the point (0,1)(0,1)(0,1).

Chords and loci built on the same ideas

Step 8: Expand and tidy.

Step 5: Expand.

Step 8: Take the midpoint of RSRSRS.

Step 3: Expand and cancel.

Step 6: Substitute.

Quick recall

Family of lines: where the $\lambda$ trick starts

Step 4: Put the centre on $y=x$ .

Step 4: Substitute and divide by $1+\lambda=2$ .

Step 4: Normalise by dividing by $3$ .

Step 5: Put the centre on $x-y=3$ .

Step 2: Impose the point $(-1,0)$ .

Step 4: Multiply through by $2$ .

Step 5: Use the point $(0,1)$ .

Step 8: Take the midpoint of $RS$ .