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

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

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

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x^{2} + y^{2} = 6 a^{2}

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

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Solution

The correct option is A $x^{2} + y^{2} = 2 a^{2}$
$S : x^{2} + y^{2} = a^{2}$
Let $P (h, k)$ be th point of intersection of perpendicular tangents to circle $S$
Tangent at $P$ , is
$T : x h + y k - a^{2} = 0$
Pair of tangent: $S S_{1} = T^{2}$
$(x^{2} + y^{2} - a^{2}) (h^{2} + k^{2} - a^{2}) = (x h + y k - a^{2})^{2}$
$\Rightarrow (k^{2} - a^{2}) x^{2} + (h^{2} - a^{2}) y^{2} - 2 h k x y + 2 a^{2} k y + 2 h a^{2} x - a^{2} (h^{2} + k^{2}) = 0$
Angle between two lines is given by $tan θ = ∣ ∣ ∣ \frac{2 \sqrt{h^{2} - a b}}{(a + b)} ∣ ∣ ∣$
$∵ θ = 90^{0}$
Coefficient of $x^{2}$ + coefficient of $y^{2} = 0$
$k^{2} - a^{2} + h^{2} - a^{2} = 0$
$\Rightarrow h^{2} + k^{2} = 2 a^{2}$
$\Rightarrow x^{2} + y^{2} = 2 a^{2}$
Locus of $(h, k)$ is a circle $x^{2} + y^{2} = 2 a^{2}$

Hence, option A.

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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} = 2 a^{2}

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

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