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Question

The line x+2y+a=0 intersects the circle x2+y2−4=0 at two distinct points A and B. Another line 12x−6y−41=0 intersects the circle x2+y2−4x−2y+1=0 at two distinct points C and D.
The equation of the circle passing through the points A,B,C and D is

A
5x2+5y28x16y36=0
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B
5x2+5y2+8x16y36=0
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C
5x2+5y2=8x+16y36=0
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D
5x2+5y28x16y+36=0
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Solution

The correct option is A 5x2+5y28x16y36=0
L1=x2+y24=0
L2=x2+y24x2y+1=0
Radical axis of C1 and C2 is :
4x+2y=5-------------(i)
4x+2y5=0-----------(ii)
x+2y+a=0-------------(iii)
If P,Q,R,S are concylic then (i),(ii),(iii) are concurvet
∣ ∣42512a12641∣ ∣=0
4(82+6a)2(4112a)5(624)=0
48a=82×3150
48a=96
a=2
The equation of finally of circle passing them PQRS is
L1+λL1=0
x2+y24+λ(x+2y+2)=0-------------(IV)
L2+λL2=0
x2+y24x2y+1+μ(12x6y41)=0------------(V)
on comparing (IV) and (V)
λ=12μ4
2λ4=141μ
65μ=13
μ=15
λ=1254
λ=85
Hence equation of circle is
x2+y24×85(x+2y+2)=0
5x2+5y28x16y36=0
Hence, option (A) is correct.



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