<s>[INST] How to find the equation of a line, tangent to a circle, that passes through a given external point I am given the equation of a circle: $(x + 2)^2 + (y + 7)^2 = 25$. The radius is $5$. Center of the circle: $(-2, -7)$.
Two lines tangent to this circle pass through point $(4, -3)$, which is outside of said circle. How would I go about finding one of the equations of the lines tangent to the circle?
I haven't started calculus, so I ask for advice fitting for someone starting a topic like this. 
 [/INST] We can use the formula for the distance from a point to a line. The distance between a line $ax + by + c = 0$ and a point $(x_0, y_0)$ is
$$\left| \frac{a x_0 + b y_0 + c}{\sqrt{a^2+b^2}} \right|.$$
The line of tangency that passes the point $(4, -3)$ has equation $y+3 = m(x-4)$ for some slope $m$. So we just need to find $m$. Note that this equation can be written as
$$mx - y -4m -3 = 0.$$
The distance between the center of the circle $(-2, -7)$ and the tangent line $mx - y -4m -3 = 0$ can be found from the formula above. This distance also equals to the radius of the circle, which is $5$. Hence
$$\left| \frac{m(-2) + (-1)(-7) -4m -3}{\sqrt{m^2 + 1}} \right| = 5.$$
Rearranging we have
$$|-6m + 4| = 5 \sqrt{m^2+1}. $$
Squaring and rearranging, we get the quadratic equation
$$11m^2 - 48m - 9 = 0$$
with solutions
$$m = \frac{24\pm15\sqrt{3}}{11}.$$
Therefore, the required equations for the tangent lines are
$$y+3 = \frac{24+15\sqrt{3}}{11}\cdot(x-4)$$
and
$$y+3 = \frac{24-15\sqrt{3}}{11}\cdot(x-4).$$
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