$x^2+y^2=5^2$ and point $(-4,3)$. Find the equations of the tangents to the circle and the point.
This question came up in class and we were unsure of how to do it. Our class spent a good 20 minutes tinkering but we ended up with a hideous quadratic. We were told that you don't need a calculator for this question, so we thought there must have been an easier way.
Does anyone have any ideas?
Is there a typo in the question?
1. If YES and if the equation of the circle should read $x^2+y^2=5^2$, then the solution is much simpler, and would be as follows:
Circle has centre $O(0,0)$ and radius $5$.
Point $P(-4,3)$ lies on the circle.
Gradient of $OP$ is $m=-\frac34$.
Tangent to circle at $P$ is perpendicular to OP, and has gradient $m'=-\frac1m=\frac43$.
Hence equation of tangent is:
$$\begin{align}y-3&=m'(x-(-4))\\
y-3&=\frac43(x+4)\\
y&=\frac43x+\frac{25}3\qquad \blacksquare \end{align}$$
2. If NO, then the solution is more messy, and it is then assumed that the question is meant to read the "tangents to the circle which passes through the point $(-4,3)$" rather than "tangents to the circle and the point".
Equation of line passing through $P(-4,3)$ with gradient $m$ is $$y-3=m(x+4)\\ y=mx+(4m+3)$$ At intersection with circle, $$x^2+(mx+(4m+3))^2=5\\ (1+m^2)x^2+2m(4m+3)x+(4m+3)^2-5=0\\$$
For tangency, $$\begin{align} [2m(4m+3)]^2&=4(1+m^2)[(4m+3)^2-5]\\ (4m+3)^2-5(1+m^2)&=0\\ 11m^2+24m+4&=0\\ (m+2)(11m+2)&=0\\ m&=-2,-\frac2{11} \end{align}$$
Hence equations of tangents are:
$$y=-2x-5$$ and $$y=-\frac2{11}x+\frac{25}{11}$$
Interestingly enough, the point $(-4, 3)$ lies on the circle:
$$(-4)^2 + 3^2 = 5^2$$
If we need to find the tangent line at a certain point, we need two things, a slope, and a point.
From the problem we already know that the line passes through $(-4, 3)$, so that is our point.
Next we need our slope. Since the line is tangent to the circle, it has the same of the circle at that tangent point.
Therefore, we must find the derivative implicitly:
$$x^2+y^2 = 25$$
$$2x + 2yy' = 0$$
$$y' = \frac{-2x}{2y} = \frac{-x}{y}$$
Substituting the point in:
$$y' = \frac{-(-4)}{3} = \frac{4}{3}$$
Now, putting this into point-slope form:
$$y-3 = \frac{4}{3}(x+4)$$
$$y = \frac{4}{3}x + \frac{16}{3} + 3 = \frac{4}{3}x + \frac{25}{3}$$
You can also parametrize the circle by $$\rho(t)=(5\cos(t),5\sin t)$$ then $$\rho'(t)=(-5\sin(t), 5\cos(t))$$ and so $$\frac{y'(t)}{x'(t)}=\frac{5\cos(t)}{-5\sin(t)}=-\frac{\cos(t)}{\sin(t)}=\frac{4}{3}$$
the last equality come from $$\begin{cases}-4=5\cos(t)\\ 3=5\sin(t)\end{cases}.$$
And so $$y=\frac{4}{3}x+b.$$ Moreover, you know that $$3=\frac{4}{3}\cdot (-4)+b\implies b=\frac{25}{3}$$ what conclude that $$y=\frac{4}{3}x+\frac{25}{3}.$$
