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Question

The line $$x+2y+a=0$$ intersects the circle $${x}^{2}+{y}^{2}-4=0$$ at two distinct points $$A$$ and $$B$$. Another line $$12x-6y-41=0$$ intersects the circle $${x}^{2}+{y}^{2}-4x-2y+1=0$$ 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 $$5{x}^{2}+5{y}^{2}-8x-16y-36=0$$
$${L_1} = {x^2} + {y^2} - 4 = 0$$
$${L_2} = {x^2} + {y^2} - 4x - 2y + 1 = 0$$
Radical axis of $${C_1}$$ and $${C_2}$$ is $$:$$
$$4x+2y=5$$-------------$$(i)$$
$$4x+2y-5=0$$-----------$$(ii)$$
$$x+2y+a=0$$-------------$$(iii)$$
If $$P,Q,R,S$$ are concylic then $$(i), (ii), (iii)$$ are concurvet 
$$\left| \begin{matrix} 4\, \, \, \, \, \, \, \, \, \, 2\, \, \, \, \, \, \, \, \, \, -5 \\ 1\, \, \, \, \, \, \, \, \, \, \, 2\, \, \, \, \, \, \, \, \, \, \, \, a \\ 12\, \, \, \, \, \, \, -6\, \, \, \, \, \, \, -41 \\  \end{matrix} \right| =0$$
$$4\left( { - 82 + 6a} \right) - 2\left( { - 41 - 12a} \right) - 5\left( { - 6 - 24} \right) = 0$$
$$48a = 82 \times 3 - 150$$
$$48a=96$$
$$a=2$$
The equation of finally of circle passing them $$PQRS$$ is 
$${L_1} + \lambda {L_1} = 0$$
$${x^2} + {y^2} - 4 + \lambda \left( {x + 2y + 2} \right) = 0$$-------------$$(IV)$$
$${L_2} + \lambda {L_2} = 0$$
$${x^2} + {y^2} - 4x - 2y + 1 + \mu \left( {12x - 6y - 41} \right) = 0$$------------$$(V)$$
on comparing $$(IV)$$ and $$(V)$$ 
$$\lambda  = 12\mu  - 4$$
$$2\lambda  - 4 = 1 - 41\mu $$
$$65\mu  = 13$$
$$\mu  = \frac{1}{5}$$
$$\lambda  = \frac{{12}}{5} - 4$$
$$\lambda  = \frac{{ - 8}}{5}$$
Hence equation of circle is 
$${x^2} + {y^2} - 4 \times \frac{{ - 8}}{5}\left( {x + 2y + 2} \right) = 0$$
$$5{x^2} + 5{y^2} - 8x - 16y - 36 = 0$$
Hence$$,$$ option $$(A)$$ is correct$$.$$



1183240_1194763_ans_ec225208d4824f4499d204e76602a3a9.PNG

Mathematics

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