Locus of the point of intersection of perpendicular tangents to the circle $x^{2} + y^{2} = 10$ is

x^{2} + y^{2} = 5

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x^{2} + y^{2} = 20

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

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x^{2} + y^{2} = 100

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Solution

The correct option is B $x^{2} + y^{2} = 20$
Equation of circle is

$x^{2} + y^{2} = 10$ …… (1)

The equation of tangent to the circle is

$y = m x + \sqrt{10} \sqrt{1 + m^{2}}$

Then, $P (h, k)$ lies on tangent then,

$(k - m h) = \sqrt{10 (1 + m^{2})}$

On squaring both side and we get,

${(k - m h)}^{2} = 10 (1 + m^{2})$

$\Rightarrow m^{2} (h^{2} - 10) - 2 m h k + k^{2} - 10 = 0$

This is the quadratic equation in $m$ .

Let,

Two roots $m_{1} a n d m_{2}$

According to given question,

$m_{1} m_{2} = - 1 (⊥ tangents)$

$\frac{k^{2} - 10}{h^{2} - 10} = - 1$

$\Rightarrow k^{2} - 10 = - h^{2} + 10$

$\Rightarrow h^{2} + k^{2} = 20$

Hence, the locus $P (h, k)$ is $x^{2} + y^{2} = 20$

Hence, this is the answer.

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

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

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

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) $x^{2} + y^{2} = 1$

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