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Circles: Entry Column Matched Value Column

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

Question
Each entry of Column-I is to be matched with its value in Column-II.
List-I
AGiven that the circumference of the circle x2+y2+6x+6y−b=0x^2+y^2+6x+6y-b=0 is bisected by the circle x2+y2−4x+2y+a=0x^2+y^2-4x+2y+a=0, the value of ∣a+b∣|a+b| is
BThe tangents to the circle x2+y2=10x^2+y^2=10 at the end points of its chord x+y=2x+y=2 meet at the point (a,b)(a,b); then the value of a2+b2a^2+b^2 is
CThe line y=2xy=2x cuts the circle x2+y2−5x=0x^2+y^2-5x=0 in the chord ABAB. Taking ABAB as diameter, the circle drawn is x2+y2−ax−by=0x^2+y^2-ax-by=0; then the value of 2a+9b2a+9b is
DThe middle points of those chords of the circle 2x2+2y2−6x+2y+1=02x^2+2y^2-6x+2y+1=0 which subtend an angle 120∘120^\circ at the centre lie on x2+y2−ax−by+2=0x^2+y^2-ax-by+2=0; then the value of a3+b3a^3+b^3 is
List-II
P2020
Q 2626
R5050
S4242
AA →\to (S); B →\to (R); C →\to (P); D →\to (Q)correct
BA →\to (P); B →\to (P); C →\to (P); D →\to (P)
CA →\to (Q); B →\to (P); C →\to (Q, R); D →\to (P)
DA →\to (P); B →\to (Q, R); C →\to (S); D →\to (S)
Solution
PART (A) Step 1: The circumference of the first circle being bisected by the second means their common chord is a diameter of the first circle, so it passes through the centre of the first circle. Step 2: The common chord is S1−S2=0S_1-S_2=0.
(x2+y2+6x+6y−b)−(x2+y2−4x+2y+a)=10x+4y−b−a=0\left(x^2+y^2+6x+6y-b\right)-\left(x^2+y^2-4x+2y+a\right) = 10x+4y-b-a = 0
 Step 3: Putting the centre (−3,−3)(-3,-3) of x2+y2+6x+6y−b=0x^2+y^2+6x+6y-b=0 in it,
10(−3)+4(−3)−b−a=0  ⟹  −30−12=a+b  ⟹  a+b=−4210(-3)+4(-3)-b-a = 0 \;\Longrightarrow\; -30-12 = a+b \;\Longrightarrow\; a+b = -42
∣a+b∣=42|a+b| = 42
 PART (B) Step 1: For x2+y2=10x^2+y^2=10 the chord of contact from (a,b)(a,b) is T=0T=0.
ax+by=10ax+by = 10
 Step 2: This is the given chord x+y=2x+y=2, so
a1=b1=102=5  ⟹  a=5, b=5\frac{a}{1} = \frac{b}{1} = \frac{10}{2} = 5 \;\Longrightarrow\; a=5,\ b=5
a2+b2=25+25=50a^2+b^2 = 25+25 = 50
 PART (C) Step 1: Every circle through AA and BB belongs to the family S+λL=0S+\lambda L=0.
x2+y2−5x+λ(2x−y)=0  ⟹  x2+y2+(2λ−5)x−λy=0x^2+y^2-5x+\lambda(2x-y) = 0 \;\Longrightarrow\; x^2+y^2+(2\lambda-5)x-\lambda y = 0
 with centre (5−2λ2, λ2)\left(\dfrac{5-2\lambda}{2},\ \dfrac{\lambda}{2}\right). Step 2: ABAB is a diameter exactly when this centre lies on the line ABAB, i.e. on y=2xy=2x.
λ2=2⋅5−2λ2=5−2λ  ⟹  λ=10−4λ  ⟹  5λ=10  ⟹  λ=2\frac{\lambda}{2} = 2\cdot\frac{5-2\lambda}{2} = 5-2\lambda \;\Longrightarrow\; \lambda = 10-4\lambda \;\Longrightarrow\; 5\lambda = 10 \;\Longrightarrow\; \lambda = 2
 Step 3:
x2+y2−5x+2(2x−y)=x2+y2−x−2y=0x^2+y^2-5x+2(2x-y) = x^2+y^2-x-2y = 0
 Comparing with x2+y2−ax−by=0x^2+y^2-ax-by=0 gives a=1a=1 and b=2b=2.
2a+9b=2(1)+9(2)=202a+9b = 2(1)+9(2) = 20
 PART (D) Step 1:
2x2+2y2−6x+2y+1=0  ⟹  x2+y2−3x+y+12=02x^2+2y^2-6x+2y+1=0 \;\Longrightarrow\; x^2+y^2-3x+y+\tfrac12 = 0
 centre O(32,−12)O\left(\dfrac32,-\dfrac12\right), radius r=94+14−12=2r=\sqrt{\dfrac94+\dfrac14-\dfrac12} = \sqrt2. Step 2: Let P(h,k)P(h,k) be the mid-point of a chord ABAB with ∠AOB=120∘\angle AOB = 120^\circ. Then OP⊥ABOP \perp AB and OPOP bisects the angle, so in the right triangle OPBOPB,
cos⁡60∘=OPOB=OP2  ⟹  OP=22=12\cos 60^\circ = \frac{OP}{OB} = \frac{OP}{\sqrt2} \;\Longrightarrow\; OP = \frac{\sqrt2}{2} = \frac{1}{\sqrt2}
 Step 3:
(h−32)2+(k+12)2=12\left(h-\frac32\right)^2+\left(k+\frac12\right)^2 = \frac12
h2−3h+94+k2+k+14=12  ⟹  h2+k2−3h+k+2=0h^2-3h+\frac94+k^2+k+\frac14 = \frac12 \;\Longrightarrow\; h^2+k^2-3h+k+2 = 0
 so the locus is x2+y2−3x+y+2=0x^2+y^2-3x+y+2=0. Comparing with x2+y2−ax−by+2=0x^2+y^2-ax-by+2=0 gives a=3a=3 and b=−1b=-1.
a3+b3=27+(−1)=26a^3+b^3 = 27+(-1) = 26
 MATCHING UP
(A)→42→(S),(B)→50→(R),(C)→20→(P),(D)→26→(Q)(A) \to 42 \to (S),\qquad (B) \to 50 \to (R),\qquad (C) \to 20 \to (P),\qquad (D) \to 26 \to (Q)
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