I am trying to find the points where lines crossing through the point (5, -2) would be tangent to the circle y^2 + x^2 = 9.
As a caveat however: this is supposed to be solvable using only college algebra. Every resource I've found thus far seems to reference calculus. Can anyone point me in the right direction?
Hint: Let $D$ be the point where the tangent line intersects the circle, let $P$ be the point $(5,-2)$ and let $O$ be the origin of the plane. The triangle $ODP$ is a right angled triangle with the right angle at $D$. You know how far away $D$ is from $O$ and you know how far away $P$ is from $O$. You can do some basic trigonometry with this information.
Let $y = mx+b$ be the equation of the tangent line. Then since $(5,-2)$ is on this line, we know that $-2 = 5m + b$ so that $\boxed{b = -2 - 5m}$.
Now consider the system of equations: $$\begin{cases} x^2 + y^2 = 9 \\ y = mx + b \end{cases}$$ Substituting, we find that: \begin{align*} x^2 + (mx+b)^2 &= 9 \\ x^2 + (m^2x^2 + 2mbx + b^2) &= 9 \\ (1 + m^2)x^2 + (2mb)x + (b^2 - 9) &= 0 \end{align*} But since the line is tangent, we know that this system must have exactly one intersection point so that the discriminant of the above quadratic equation must be zero. Thus: \begin{align*} (2mb)^2 - 4(1+m^2)(b^2 - 9) &= 0 \\ 4m^2b^2 - 4b^2 + 36 - 4m^2b^2 + 36m^2 &= 0 \\ 9m^2 - b^2 + 9 &= 0 \\ 9m^2 - (-2-5m)^2 + 9 &= 0 \\ 9m^2 - (4 + 20m + 25m^2) + 9 &= 0 \\ -16m^2 - 20m + 5 &= 0 \\ m &= \frac{-(-20) \pm \sqrt{(-20)^2 - 4(-16)(5)}}{2(-16)} \\ m &= \frac{-5 \pm 3\sqrt 5}{8} \end{align*}
Thus, the equations of the two tangent lines are: $$\begin{cases} y = \left(\dfrac{-5 - 3\sqrt 5}{8} \right)x -2 - 5\left(\dfrac{-5 - 3\sqrt 5}{8} \right) \\ y = \left(\dfrac{-5 + 3\sqrt 5}{8} \right)x -2 - 5\left(\dfrac{-5 + 3\sqrt 5}{8} \right) \end{cases}$$
Hint. The tangent line at (5,-2) is perpendicular to the radius passing through (5,-2)
