Finding the point of tangency between a curve and a line

Problem:

• a circle with radius $2$ centered at the origin, and
• a line which passes through the point $(3,0)$

are tangent to each other in the third quadrant, determine the point of tangency.

Solution:

Let $(a,b)$ be the coordinates of the point of intersection.

Because $(a,b)$ lies on the edge of the circle, it must satisfy $a^2 + b^2 = 4$. Solving for $b$, we have $b=\pm\sqrt{4-a^2}$. Since the point of tangency is in the third quadrant, its coordinates are therefore $(a, -\sqrt{4-a^2})$.

Next we calculate the distance between $(3,0)$ and $(a,-\sqrt{4-a^2})$: \begin{align*} d &= \sqrt{(3-a)^2 - (0+\sqrt{4-a^2})^2}\\ &= \sqrt{9 - 6a + a^2 + 4 - a^2}\\ &= \sqrt{13-6a} \end{align*} The tangent line, the radius of the circle, and the $x$-axis form a right triangle. Therefore, according to the Pythagorean thoerem: \begin{align*} d^2 + 2^2 &= 3^2\\ \left(\sqrt{13-6a}\right)^2 + 2^2 &= 3^2\\ 13-6a + 4 &= 9\\ 6a &= 8\\ a &= \frac{4}{3} \end{align*} Substituting into the expression for $b$, we get $$b = -\sqrt{4-a^2} = -\sqrt{4-\left(\frac{4}{3}\right)^2} = -\sqrt{\frac{36}{9}-\frac{16}{9}} = -\sqrt{\frac{20}{9}} = -\frac{2\sqrt{5}}{3}$$ Thus, the point of intersection is $$ \boxed{\left(\frac{4}{3},-\frac{2\sqrt{5}}{3}\right)}$$