There are two circles that pass through (1,9) and (8,8) and are tangent to the x-axis. Find the lengths of their radii.

Guest Dec 11, 2019

#1**0 **

The first step is to draw the perpendicular bisector of the segment AB since it would contain centers of the circles that pass through both points. If we call the centers, only one in the image, O(a, b), and the point of tangency H, then the three segments AO, BO, and HO would have to have the same length and that length is b, the radius of the circle. So we can write the following:

\((a-1)^2+(b-9)^2=(a-8)^2+(b-8)^2=b^2\)

This is a system of two equations and two unknowns that can be written as follows:

\(\frac{(a-1)^2}{9}=2b-9\)

\(\frac{(a-8)^2}{8}=2b-8\)

which implies that

\(\frac{(a-1)^2}{9}+9=\frac{(a-8)^2}{8}+8\)

Simplifying, we end up with the quadratic equation \(a^2-128a+496= (a-4)(a-124)=0\) that has solutions

\(a=124\) and \(a=4\). These values of \(a\) result in two values for \(b\) giving the radii of the circles:

\(\frac{(4-1)^2}{9}=2b-9\), implying \(b=5\), and

\(\frac {(124-1)^2}{9}=2b-9\), implying \(b=845\), a rather surprising result.

Gadfly Dec 12, 2019