# Very Interesting, Very Challenging Circle Problem

• February 13th 2008, 04:13 PM
hummeth
Very Interesting, Very Challenging Circle Problem
There is a circle radius of $\sqrt {50}$
There are two line segments legnths 2 and 6 each with one end point on the circle, the other ends meeting at a 90 degree angle. What is the distance between the center of the circle and the right angle where the two segments meet?
• February 13th 2008, 04:24 PM
Gauss
Are the line segments inside or outside the circle?(Thinking)
• February 13th 2008, 05:31 PM
hummeth
Quote:

Originally Posted by Gauss
Are the line segments inside or outside the circle?(Thinking)

Inside, sorry I wasn't specific enough.
• February 14th 2008, 03:31 AM
earboth
Quote:

Originally Posted by hummeth
There is a circle radius of $\sqrt {50}$
There are two line segments legnths 2 and 6 each with one end point on the circle, the other ends meeting at a 90 degree angle. What is the distance between the center of the circle and the right angle where the two segments meet?

1. Draw a sketch.

2. You are dealing with 2 right triangles. Use Pythagorean theorem:

$\left|\begin{array}{l}y^2 + (6-x)^2 = (\sqrt{50})^2\\ (2+y)^2 + x^2 = 50\end{array} \right.$ ..... Expand the brackets, collect like terms:

$\left|\begin{array}{l}y^2 + 36-12x+x^2 = 50\\ 4+4y+y^2 + x^2 = 50\end{array} \right.$ ..... Subtract the 2nd equation from the first:

$32-4y-12x = 0~\implies~y = -3x+8$ ..... Plug in this term for y into the 1rst equation. You get a quadratic equation in x:

$10x^2-60x+50 = 0~\implies~x = 5~\vee~x=1$ ..... These x-values correspond with the y-values y = -3 (rejected) or y = 5.

Therefore the distance from the point of intersection to the center of the circle is:

$d = \sqrt{5^2+1^2}=\sqrt{26}\approx 5.099$
• February 14th 2008, 04:20 AM
hummeth
That's perfect! I figured it out last night with a more algebraic approach. I put the circle on a coordinate plane centerting the circle on the vertex. Then, I set the right angle as point (a,b) and made the two ling segments vertical and horizontal lines on the coordinate plane. The points on the circle were then (a,y) and (x,b) Then I solved these systems.
x^2+y^2=50
a^2+y^2=50
x^2+b^2=50
y-b=6
x-a=2
After some algebra, I got the same answer, roughly 5.1