Tangent lines to circle $ x^2+y^2=100 $ passing point (14, 2)

Question

I'm asked to give tangent lines to $ x^2+y^2=100 $, so that both tangent lines go through the point $ (14,2)$.

Implicit differentiation gives:

$dy/dx=-x/y$

While graphing I noted that the circle , doesn't go through (14,2).

I constructed the following tangent line to the point (14,2):

$ y-2=-7(x-14) $

but it's not tangent to the circle.

I'm a bit stuck, need some help :D

Answer 1

Let the tangent lines be $x+ay-2a-14=0$. Use the distance formula between the lines and the origin,

$$\frac{|1(0)+a(0)-2a-14|}{\sqrt{1+a^2}}=10$$

which yields $a=-\frac34,\>\frac43$. Thus, the tangent lines are

$$4x-3y-50=0,\>\>\>\>\>3x+4y-50=0$$

Answer 2

We form the system $$\begin{cases}y-2=m(x-14),\\x^2+y^2=100\end{cases}$$

or, by elimination

$$x^2+(m(x-14)+2)^2=100.$$

This equation has a double root (hence the line is tangent) when the delta cancels, i.e. when

$$12m^2-7m-12=0.$$

Answer 3

Similar question I have seen on other forum. Link given https://www.mathsdiscussion.com/forum/topic/tanget-to-circle/?part=1#postid-38

In your question a=10 and satisfy point (14,2) on tangent you get quadratic in m find m.

Answer 4

Following your idea, the line slope is:

$$\frac{dy}{dx}=-\frac{x_0}{y_0}$$

where $(x_0,y_0)$ is a point of the circle. So, you can't use $(x_0,y_0)=(14,2)$ because it is not a point of the circle.

Considering that, the best idea is to use @Yves's approach.