Two perpendicular chords AB and CD intersect at P. If PA = 4, PB = 10, and CD = 13, calculate the length of P to the center of the circle
+1164 There’s probably more than one way to solve this. Here’s how I would go about it.
From the intersecting chords theorem, you know that CPxDP = 4x10. You also know that CP+DP=13 (=CD). You now have two equations involving CP and DP.
Combining the two equations gives you a quadratic expression which when solved gives you CP= 5 and DP=8.
The distance from the midpoint of CD to P can be found to be 1.5.
The distance from the midpoint of AB to P can be found to be 3.0.
The distance of P to the center of the circle is the same as the diagonal of a rectangle of sides 1.5 and 3.0 which can be evaluated from the Pythagorean theorem (approximately 3.354).
quora
+2668 Let \(CP = x\) and \(DP = y\) and let the center of the circle be o.
We have the following system from the intersecting chord theorem:
\(xy = 40 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (i)\)
\(x + y = 13 \ \ \ \ \ \ \ \ \ (ii)\)
Solving for x and y, we find that \(x = 8\) and \(y = 5\).
P is 3 units away from the midpoint of AB and 1.5 units away from the midpoint of CD.
This means that the distance is \(\sqrt{3^2 + 1.5^2} = \sqrt{11.25} = \color{brown}\boxed{3 \sqrt 5 \over 2}\)
Here's a graph: 
