1966 AHSME Problems/Problem 8

Problem

The length of the common chord of two intersecting circles is $16$ feet. If the radii are $10$ feet and $17$ feet, a possible value for the distance between the centers of the circles, expressed in feet, is:

$\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}$

Solution

File:1966 AHSME-8.jpg Let $O$ be the center of the circle of radius $10$ and $P$ be the center of the circle of radius $17$ . Chord $\overline{AB} = 16$ feet. $\overline{OA} = \overline{OB} = 10$ feet, since they are radii of a circle. Hence, $\triangle OAB$ is isoceles with base $AB$ . The height of $\triangle OAB$ from $O$ to $AB$ is $\sqrt {\overline{OB}^2 - (\frac{\overline{AB}}{2})^2} = \sqrt {10^2 - (\frac{16}{2})^2} = \sqrt {100 - 8^2} = \sqrt {100 - 64} = \sqrt {36} = 6$

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1966 AHSME Problems/Problem 8

Problem

Solution