Question

If two perpendicular tangents can be drawn from the origin to the circle $x^2-6x+y^2-2py+17=0$, then find the value of $|p|$.

Answer 1

The equation of given circle is $x^2-6x+y^2-2py+17=0$ or ${(x-3)}^2+{(y-p)}^2=(p^2-8)$ (1)

Also$(0,0)$ lies outside the circle.

Equation of director circle of $S-0$ will be

${(x-3)}^2+{(y-p)}^2=2(p^2-8)$ (2)

Tangents drawn from $(0,0)$ to circle (i) are perpendicular to each other

$\therefore (0,0)$ must lie on director circle.

$\therefore {(0-3)}^2+{(0-p)}^2=2(p^2-8)$

$\Rightarrow p^2=25$

$\Rightarrow p=\pm 5$

$\therefore |p|=5$