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# Circles

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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

Jul 29, 2022

#1
+1132
+6

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

Jul 29, 2022
edited by nerdiest  Jul 29, 2022
#3
+2444
0

you beat me to it....

just barely....

I'll beat you next time lol

BuilderBoi  Jul 29, 2022
#4
+1132
+6

LOL

I saw you doing your answer, and I was like, I gotta beat BuilderBoi

nerdiest  Jul 29, 2022
#2
+2444
0

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:

Jul 29, 2022
#5
+124594
+1

Very nice , guys   !!!!!

Jul 29, 2022