Question

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

• A. $0$
• B. $3$
• C. $5$
• D. $17$

Answer 1

The correct option is C $5$

To a circle $(x-h{)}^2+(y-k{)}^2=r^2$, perpendicular tangents can be drawn from a point on the circle which satisfies the equation $(x-h{)}^2+(y-k{)}^2=2r^2$

Here, the circle's equation can be written as $(x-3{)}^2+(y-p{)}^2=p^2-8$

Thus, the locus of points from where perpendicular tangents can be drawn to this circle is given by $(x-3{)}^2+(y-p{)}^2=2(p^2-8)$

Since origin lies on this locus, $(0-3{)}^2+(0-p{)}^2=2(p^2-8)$

$\therefore 9+p^2=2p^2-16$

$\therefore p^2=25$

$\therefore |p|=5$