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

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$$.$$Mathematics

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