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Question

Suppose $$ax + by + c = 0$$, where a, b, c are in A. P. be normal to a family of circles. The equation of the circle of the family intersects the circle $$x^2 + y^2 - 4x - 4y - 1 = 0$$ orthogonally is


A
x2+y22x+4y3=0
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B
x2+y2+2x4y3=0
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C
x2+y22x+4y5=0
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D
x2+y22x4y+3=0
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Solution

The correct option is A $$x^2 + y^2 - 2x + 4y - 3 = 0$$
a, b, c are in A. P., so $$ax + by + c = 0$$ represents a family of lines passing through the point $$(1, -2)$$.
So, the family of circles (concentric) will be given by $$x^2 + y^2 - 2x + 4y + c = 0$$. It intersects given circle orthogonally.
$$\Rightarrow  2( -1 \times -2) + (2 \times -2) = -1 + c \Rightarrow c = - 3$$

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