JAMB 1993 · UME · Q34

Circle geometry

In the figure, the line segment $ST$ is a tangent to the two circles at $S$ and $T$ . $O$ and $Q$ are the centres of the circles with $OS = 5$ cm, $QT = 2$ cm and $OQ = 14$ cm. Find $ST$ .

Worked solution (try it first)

Radii meet a tangent at $90^\circ$ , so $OS \perp ST$ and $QT \perp ST$ .
The tangent crosses between the circles, so the radii point opposite ways.
Slide $QT$ along to the end of $OS$ : this makes a right-angled triangle with hypotenuse $OQ = 14$ , one side $5 + 2 = 7$ and the other side equal to $ST$ .
Pythagoras: $ST^2 = 14^2 - 7^2 = 147$ , so $ST = \sqrt{147} = 7\sqrt3$ cm, option A.

Report a problem with this question