Two circles with centres \(A\) and \(B\), each with a radius of \(3\), are tangent to each other at \(S\). A straight line is drawn through \(A\), \(S\) and \(B\), meeting the circle with centre \(A\) at \(Q\), \(Q\neq S\), and the circle with centre \(B\) at \(R\), \(R\neq S\). Point \(P\) is then drawn so that \(PQ\) and \(PR\) are each tangent to the circle with centre \(A\).
Determine the length of \(PQ\).
Note: For this problem, you may want to use the following known results about circles:
If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangency.
A line drawn from the centre of a circle perpendicular to a tangent line meets the tangent line at the point of tangency.
Since each circle has a radius of \(3\), \(AQ=AS=BS=BR=3\). From Result 1 given after the problem, since \(PQ\) is tangent to the circle with centre \(A\), it follows that \(PQ \perp QR\).
Let \(T\) be the point of tangency of the line \(PR\). Then \(AT \perp PR\). Since \(AT\) is a radius of the circle, \(AT=3\). Since \(\triangle ATR\) is a right-angled triangle, \[TR^2 = AR^2-AT^2=9^2-3^2=72\] Thus, \(TR=\sqrt{72}=6\sqrt{2}\), since \(TR>0\).
We will now proceed with three different approaches.
Solution 1
In this solution we use trigonometry. Since \(PQ \perp QR\) and \(AT \perp PR\), \(\triangle PQR\) and \(\triangle ATR\) are right-angled triangles. In \(\triangle PQR\), \(\tan{R}=\dfrac{PQ}{QR}=\dfrac{PQ}{12}\). Similarly, in \(\triangle ATR\), \(\tan{R}=\dfrac{AT}{TR}=\dfrac{3}{6\sqrt{2}}\). Thus, \[\begin{aligned} \frac{PQ}{12} &= \frac{3}{6\sqrt{2}}\\ PQ &= \frac{3}{6\sqrt{2}} \times 12 = \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \end{aligned}\] Therefore, the length of \(PQ\) is \(3\sqrt{2}\).
Solution 2
In this solution we use similar triangles. Since \(\angle PQR = \angle ATR = 90\degree\) and \(\angle PRQ = \angle ART\), it follows that \(\triangle PQR \sim \triangle ATR\). Thus, \[\begin{aligned} \frac{PQ}{QR} &= \frac{AT}{TR}\\ \frac{PQ}{12} &= \frac{3}{6\sqrt{2}}\\ PQ &= \frac{3}{6\sqrt{2}} \times 12 = \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \end{aligned}\] Therefore, the length of \(PQ\) is \(3\sqrt{2}\).
Notice that while the initial approach is different, this solution is quite similar to Solution 1.
Solution 3
In this solution we use the Pythagorean Theorem. Draw line segment \(AP\). Then \(\triangle PQA\) and \(\triangle PTA\) are right-angled triangles.
Thus, \(AP^2=PQ^2+AQ^2\) and \(AP^2=PT^2+AT^2\). Since \(AQ\) and \(AT\) are both radii, \(AQ=AT\). Thus, \(PQ=PT\). Let \(x=PQ=PT\).
Since \(PQ \perp QR\), \(\triangle PQR\) is right-angled with \(PQ=x\), \(QR=12\), and \(PR=x+6\sqrt{2}\). Thus, \[\begin{aligned} PR^2 &= PQ^2 + QR^2\\ (x + 6\sqrt{2})^2 &= x^2 + 12^2\\ x^2+(12\sqrt{2})x + 72 &= x^2 + 144\\ (12\sqrt{2})x &= 72\\ x &= \frac{72}{12\sqrt{2}} = \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \end{aligned}\] Therefore, the length of \(PQ\) is \(3\sqrt{2}\).