Suppose $a x + b y + c = 0$ , 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 $x^{2} + y^{2} - 4 x - 4 y - 1 = 0$ orthogonally is

x^{2} + y^{2} - 2 x + 4 y - 3 = 0

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x^{2} + y^{2} + 2 x - 4 y - 3 = 0

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x^{2} + y^{2} - 2 x + 4 y - 5 = 0

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x^{2} + y^{2} - 2 x - 4 y + 3 = 0

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Solution

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

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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 x2+y2−4x−4y−1=0 orthogonally is

Suppose $a x + b y + c = 0$ , 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 $x^{2} + y^{2} - 4 x - 4 y - 1 = 0$ orthogonally is