Find the coordinates of T (point where tangent touches the circle).

Question

Given that P (a point that lies on the tangent) $= (6,-6)$ and the equation of the circle is $(x+5)^2 + (y-4)^2 = 25$.

I'm unsure about my answer to this question and would like to know what answers you guys got. It says that there could be more than one possible pair of coordinates for T. Also, why is it not necessary to know the position of T to answer the question?

My answers to this question are $T = (-23/2, 761/24)$ or $T= (-13/6, 149/16)$.

Any help would be appreciated. Thank you!

Answer 1

Note that the given point $(6,-6)$, the center $(-5,4)$ and the tangent point $T(p,q)$ form a right triangle, which via Pythagorean theorem leads to

$$(p-6)^2+(q+6)^2+25= 11^2+10^2$$

Also, $(p,q)$ is on the circle

$$(p+5)^2+(q-4)^2=25$$

Solve the joint equations to obtain the two tangent points

$$(p,q)=(-\frac{90}{13},-\frac8{13}),\> (-\frac{10}{17},\frac{108}{17})$$

Answer 2

Here's one possible approach to this problem. We can use the property that any tangent line to a circle has precisely one common point with that circle.

So, from the given point $P(6,-6)$ we need to draw a line tangent to the given circle. Any non-vertical line thru this point can be written in the form $y+6=m(x-6)$, or equivalently $y=mx-6m-6$, where the slope $m$ is unknown. So now we need to solve the system of equations for the points of intersection of this line with the given circle — and we want this system to have a unique solution. So plug in $y=mx-6m-6$ into the equation of the circle to get a quadratic equation for $x$, and make sure that this equation has one root — that will tell you what $m$ is.