Chords and radii?

Geometry Level 4

Two perpendicular chords of a circle $A_{1}A_{4}$ and $A_{2}A_{3}$ intersect at a point $P$ . If for $i=1, 2, 3$ , the length of $PA_{i}=2^{i}$ , the radius of the circle can be expressed as $\sqrt{a}$ where $a$ is a positive square-free integer. Find $a$ .

The answer is 85.

1 solution

Saket Sharma
Sep 27, 2014

Don't know how to create a diagram here. But, I shall try to explain.

Let $C$ be the centre and $r$ be the radius of the circle. Let $M$ and $N$ be feet of perpendiculars from $C$ on $A_1 A_4$ and $A_2 A_3$ respectively.

Now, notice that $NA_2 = NA_3 = \frac{(8+4)}{2} = 6$

and $CM = NP = NA_2 - PA_2 = 6 - 4 = 2$ .

Further, (if you draw the diagram) you can see that $CN = MP$ .

Now, $CN = \sqrt{r^{2} - 36}$

and, $MP = MA_1 - PA_1 = \sqrt{r^{2} - 4} - 2$

Thus, $\sqrt{r^{2} - 36} = \sqrt{r^{2} - 4} - 2$

Squaring on both sides once and solving we get, $\sqrt{r^{2} - 4} = 9$

whence, $a = r^{2} = 85$

Chords and radii?

The answer is 85.

1 solution

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