The locus of point of intersection of perpendicular tangents to the circle $x^{2} + y^{2} = 8$ is

x^{2} + y^{2} = 4

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x^{2} + y^{2} = 32

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x^{2} + y^{2} = 16

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x^{2} + y^{2} = 2

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Solution

The correct option is C $x^{2} + y^{2} = 16$
The locus of point of intersection of perpendicular tangents to the circle $x^{2} + y^{2} = 8$ is a directer circle
which is $x^{2} + y^{2} = 2 (8) = 16$
Ans: C

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x^{2} + y^{2} = a^{2}

The locus of the point of intersection of perpendicular tangents to the circles $x^{2} + y^{2} = a^{2} a n d x^{2} + y^{2} = b^{2} i s$

x^{2} + y^{2} = 16

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x^{2} + y^{2} = 10

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