Suppose two perpendicular tangents can be dra | vedclass

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Question

(c)

We have,

Equation of circle

$x^2+y^2-6 x-2 p y+17 =0$

$\Rightarrow \quad(x-3)^2+(y-p)^2 =p^2-8$

Given two perpendicular tangents dr

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(c)

We have,

Equation of circle

$x^2+y^2-6 x-2 p y+17 =0$

$\Rightarrow \quad(x-3)^2+(y-p)^2 =p^2-8$

Given two perpendicular tangents drawn from origin to the circle.

$	herefore$ Locus is director circle of the given circle.

$	herefore$ Equation of director circle is

$(x-3)^2+(y-p)^2=2\left(p^2-8ight)$

This passes through $(0,0)$.

$	herefore 9+p^2=2 p^2-16$

$\Rightarrow p^2=25 \Rightarrow p=\pm 5$

$	herefore |p|=|\pm 5|=5$

Suppose two perpendicular tangents can be drawn from the origin to the circle $x^2+y^2-6 x-2 p y+17=0$, for some real $p$. Then, $|p|$ is equal to

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