Two perpendicular tangents to the circle x^2 | vedclass

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Question

(A) Let the point $P$ be $(h, k)$.Since the tangents are perpendicular,the angle between them is $90^{\circ}$.The locus of the intersection of two per

Vedclass · Accepted Answer

$x^2 + y^2 = 2a^2$

Vedclass · Answer

(A) Let the point $P$ be $(h, k)$.Since the tangents are perpendicular,the angle between them is $90^{\circ}$.The locus of the intersection of two perpendicular tangents to a circle is known as the director circle.For a circle $x^2 + y^2 = a^2$,the radius is $a$ and the center is $(0, 0)$.The distance from the center $(0, 0)$ to the point $P(h, k)$ is $\sqrt{h^2 + k^2}$.In the right-angled triangle formed by the center,the point of contact,and $P$,the distance $PC$ is given by $\sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$.Thus,$\sqrt{h^2 + k^2} = a\sqrt{2}$.Squaring both sides,we get $h^2 + k^2 = 2a^2$.Replacing $(h, k)$ with $(x, y)$,the equation of the locus is $x^2 + y^2 = 2a^2$.

Two perpendicular tangents to the circle $x^2 + y^2 = a^2$ meet at point $P$. The equation of the locus of $P$ is:

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