   Chapter 6.CR, Problem 3CR ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
1 views

# Two circles intersect and have a common chord 10 in. long. The radius of one circle is 13 in. long and the centers of the circles are 16 in. apart. Find the radius of the other circle.

To determine

To find: The radius length of the second circle.

Explanation

Theorem:

1) A radius that is perpendicular to a chord bisects the chord.

2) If a line through the centre of a circle bisects a chord other than a diameter, then it is perpendicular to the chord.

Calculation:

Let the centers of the two circles be A and B.

And also let the common chord be PQ.

Given that the length of the radius of one circle with center A is AP = 13 in.

The distance between the centers of the circles = AB = 16 cm

The length of the common chord = PQ = 10 in.

Let the distance from the center of the circle with center A to the chord OA = x in.

By theorem, OAPQ, OBPQ.

So, ◿OAP is a right triangle.

By theorem, OQ=PO or PQ = 2PO = 2QO.

So, PO=12PQ.

So, OP=OQ=1210=5 in.

In right triangle OAP, by Pythagoras theorem

AP2=OP2+OA2

Let AP= 13 in, OP = 5 in, OA = x in

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