Condition for the Line y = mx + c to be a Tangent to the Circle x²+y²=a², and Finding the Point of Contact

Let the circle be $x^2+y^2=a^2$ (centre the origin, radius $a$) and the line $y=mx+c$.

1. Condition to be a tangent. The perpendicular distance from $(0,0)$ to $mx-y+c=0$ is $\dfrac{|c|}{\sqrt{1+m^2}}$. The line touches the circle exactly when this distance equals the radius $a$:

   $\frac{|c|}{\sqrt{1+m^2}}=a \iff c^2=a^2(1+m^2).$

   So $y=mx\pm a\sqrt{1+m^2}$ is a tangent to $x^2+y^2=a^2$ for every slope $m$ — there are always two parallel tangents of a given slope, one on each side of the circle.
2. Point of contact. Let $(x_1,y_1)$ be the point of contact. Since it lies on the line, $y_1=mx_1+c$. The tangent at $(x_1,y_1)$ (from §5.2.2, general-form tangent with $g=f=0,c\to -a^2$) is $xx_1+yy_1=a^2$, i.e. $y=-\frac{x_1}{y_1}x+\frac{a^2}{y_1}$. Matching this with $y=mx+c$ (same line, so coefficients are proportional) gives $\frac{x_1}{-m}=\frac{y_1}{1}=\frac{a^2}{c}$, so $y_1=\frac{a^2}c$ and $x_1=-\frac{a^2m}c$; using $c=\pm a\sqrt{1+m^2}$, the point of contact is $\left(\mp\dfrac{am}{\sqrt{1+m^2}},\ \pm\dfrac{a}{\sqrt{1+m^2}}\right)$. …