Locus of the point of intersection of perpendicular tangents to the circle $x^{2} + y^{2} = 16$ is

x^{2} + y^{2} = 8

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

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x^{2} + y^{2} = 64

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

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Solution

The correct option is B $x^{2} + y^{2} = 32$
We know that, if two perpendicular tangents to the circle $x^{2} + y^{2} = a^{2}$ meet at P, then the point P lies on a director circle.
$∴$ Required locus is $x^{2} + y^{2} = 32$

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

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List- I	List- II
A. The locus of the point of intersection of the perpendicular tangents of $x^{2} + y^{2} = a^{2}$	1) $x^{2} + y^{2} = 5$ of
B. The locus of the mid-point of the chord of the circle $x^{2} + y^{2} = 8$ which is $\sqrt{2}$ units from origin is	2) $x^{2} + y^{2} = a^{2} {sec}^{2} α$
C. The locus of the point of intersection of the lines $x = \frac{1 - t^{2}}{1 + t^{2}}$ and $y = \frac{2 t}{1 + t^{2}}$	3) $x^{2} + y^{2} = 2 a^{2}$
D. The locus of the point whose chord of contact w.r.t $x^{2} + y^{2} = a^{2}$ making an angle '2 $α$ ' at the center is	4) $x^{2} + y^{2} = 2$ 5) $x^{2} + y^{2} = 1$

x^{2} + y^{2} = 9

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

x^{2} + y^{2} = 9

x^{2} + y^{2} = 9

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