1957 AMC 12 Problem 46

Question

Two perpendicular chords intersect in a circle. The segments of one chord are

3

and

4;

the segments of the other are

6

and

2.

Then the diameter of the circle is:

\sqrt{89}

\sqrt{56}

\sqrt{61}

\sqrt{75}

\sqrt{65}

Place the chord intersection at the origin and the perpendicular chords on the coordinate axes

The circle’s center lies on both chord perpendicular bisectors

Place the intersection at

(0,0)

with endpoints

(-3,0),

(4,0),

(0,-2),

and

(0,6).

The perpendicular bisectors of the chords are

x=\frac{1}{2}

and

y=2,

so the center is

(\frac{1}{2},2).

Using the endpoint

(4,0),

\begin{aligned} r^2 &=\left(4-\frac12\right)^2+(0-2)^2\\ &=\frac{65}{4}. \end{aligned}

Therefore the diameter is

2r=\sqrt{65}.

Thus, the correct answer is E.

Problem 46 in other years