CirclesmediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Chord Length of a Circle: (AB)^2 = 18 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let a circle pass through the origin and its centre be the point of intersection of the two mutually perpendicular lines x+(k−1)y+3=0x+(k-1)y+3=0 and 2x+k2y−4=02x+k^2y-4=0. If the line x−y+2=0x-y+2=0 intersects the circle at the points AA and BB, then (AB)2(AB)^2 is equal to
A1010
B2727
C1818correct
D3434
Solution
Step 1: The lines are perpendicular, so the product of their slopes is −1-1:
(11−k)(2k2)=1 ⇒ 2=k2−k3 ⇒ k3−k2+2=0 ⇒ k=−1.\left(\frac{1}{1-k}\right)\left(\frac{2}{k^2}\right)=1\ \Rightarrow\ 2=k^2-k^3\ \Rightarrow\ k^3-k^2+2=0\ \Rightarrow\ k=-1.
 Step 2: Put k=−1k=-1 in the two lines: x−2y+3=0x-2y+3=0 and 2x+y−4=02x+y-4=0. Solving:
−5y+10=0 ⇒ y=2,x=1.-5y+10=0\ \Rightarrow\ y=2,\qquad x=1.
 So the centre is (1,2)(1,2). Step 3: The circle passes through the origin, so its radius is
r=12+22=5,r=\sqrt{1^2+2^2}=\sqrt5,
 and the circle is (x−1)2+(y−2)2=5(x-1)^2+(y-2)^2=5. Step 4: Perpendicular distance from centre (1,2)(1,2) to the chord x−y+2=0x-y+2=0:
p=∣1−2+2∣12+12=12.p=\frac{|1-2+2|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt2}.
 Step 5: Half-chord length:
ℓ=r2−p2=5−12=92=32.\ell=\sqrt{r^2-p^2}=\sqrt{5-\tfrac12}=\sqrt{\tfrac92}=\frac{3}{\sqrt2}.
 Step 6: Full chord AB=2ℓ=62AB=2\ell=\dfrac{6}{\sqrt2}, so
(AB)2=362=18.(AB)^2=\frac{36}{2}=18.
 Correct answer: (3)
Solution working
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