# Thread: Find the center of a circle with two tangent lines

1. ## Find the center of a circle with two tangent lines

Hello, I ran into a problem in my pre-cal textbook that want me to find the center of circle with two tangent lines. One line is x-2y+4=0 tangents at (0,2), the other is y=2x-7 tangents to the same circle at (3,-1). At this point the textbook hasn't introduced trigonometric functions as well as properties of circle so I would like an explanation without any of those terminologies. Here are some of a few hints given before the question:
- equation of a circle: x^2+y^2=r^2
- equation of the tangent line: y=mx+b
- r^2(1+m^2)=b^2 (don't really understand what that mean)
- point of tangency [(-r^2m)/b , (r^2)/b]
- the tangent line is perpendicular to the line containing the center of the circle and the point of tangency

This is my first time using the forum so if there is anything that isn't clear please let me know. Thank you very much.

2. ## Re: Find the center of a circle with two tangent lines

Find the lines perpendicular to given tangent lines passing through tangent points. Intersection S of those lines is the center of the circle, and radius is the distance between S and either of the two given tangent points.

3. ## Re: Find the center of a circle with two tangent lines

Originally Posted by dotdotdot
Hello, I ran into a problem in my pre-cal textbook that want me to find the center of circle with two tangent lines. One line is x-2y+4=0 tangents at (0,2), the other is y=2x-7 tangents to the same circle at (3,-1).
@dotdotdot, you have been given good suggestions on what needs doing.

Here are some practical approaches.
Suppose that $a\cdot b\ne0$. So that $\ell: ax+by+c=0$ is a line & $P: (p,q)$ is a point.
Then the line $bx-ay-(bp-aq)=0$ is perpendicular to $\ell$ and contains the point $P$.

Now write the two lines and find their intersection, the center of the circle.