The tangents drawn from the origin to the circle x^2 + y^2 -2rx -2hy +h^2=0 are perpendicular if?

A) $h = r$ B)$h = - r$ C) ${r}^{2} + {h}^{2} = 1$ D) ${r}^{2} + {h}^{2} = 2$

Oct 20, 2017

Both (A) and (B).

Explanation:

The center of circle ${x}^{2} + {y}^{2} - 2 r x - 2 h y + {h}^{2} = 0$ or ${\left(x - r\right)}^{2} + {\left(y - h\right)}^{2} = {r}^{2}$ is $\left(r , h\right)$ and radius is $r$.

As $x$-coordinate of center is $r$ and radius too is $r$, one of the tangent is $x = 0$, the $y$-axis and as other tangent would be parallel to $x$-axis.

Further, as we are drawing tangent from origin i.e. $\left(0 , 0\right)$, it can only be $x$-axis i.e. $y = 0$.

As such, distance from center $\left(r , h\right)$ to $x = 0$ i.e. $\pm h = r$.

Hence answer could be both (A) and (B).

As an example see the following graph.

graph{((x-5)^2+(y-5)^2-25)((x-5)^2+(y+5)^2-25)=0 [-19.92, 20.08, -9.76, 10.24]}