Question

pts on intersection of tangent of slope m on circle

Answer 1

The question is incomplete. Assuming it refers to the points of tangency for a tangent of slope $m$ on the circle $x^2+y^2=r^2$, the points are $\left( \mp \frac{mr}{\sqrt{1+m^2}}, \pm \frac{r}{\sqrt{1+m^2}} \right)$.

Answer 2

For the circle $x^2 + y^2 = r^2$, the equation of a tangent with slope $m$ is $y = mx \pm r\sqrt{1+m^2}$. The point of tangency $(x_1, y_1)$ satisfies both the circle equation and the tangent equation. The slope of the radius to the point of tangency is $-\frac{x_1}{y_1}$. Since the tangent is perpendicular to the radius at the point of tangency, the slope of the tangent is $m = -\frac{y_1}{x_1}$, so $y_1 = -mx_1$. Substituting this into $x_1^2 + y_1^2 = r^2$, we get $x_1^2 + (-mx_1)^2 = r^2$, which simplifies to $x_1^2(1+m^2) = r^2$, so $x_1 = \pm \frac{r}{\sqrt{1+m^2}}$. Then $y_1 = -m \left( \pm \frac{r}{\sqrt{1+m^2}} \right) = \mp \frac{mr}{\sqrt{1+m^2}}$. Thus, the points of tangency are $\left( \pm \frac{r}{\sqrt{1+m^2}}, \mp \frac{mr}{\sqrt{1+m^2}} \right)$.