Answer
If the circle has radius $r$, equation is
$x^{2}+y^{2}$ = $r^{2}$
Differentiating
$2x+2yy'$ = $0$
$y'$ = $-\frac{x}{y}$
so slope of the tangent line at $P(x_{0},y_{0})$ is $-\frac{x_{0}}{y_{0}}$
The negative reciprocal of that slope is $\frac{y_{0}}{x_{0}}$, which is slope of OP
so the tangent line at P is perpendicular to the radius OP
Work Step by Step
If the circle has radius $r$, equation is
$x^{2}+y^{2}$ = $r^{2}$
Differentiating
$2x+2yy'$ = $0$
$y'$ = $-\frac{x}{y}$
so slope of the tangent line at $P(x_{0},y_{0})$ is $-\frac{x_{0}}{y_{0}}$
The negative reciprocal of that slope is $\frac{y_{0}}{x_{0}}$, which is slope of OP
so the tangent line at P is perpendicular to the radius OP
