Question

PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

Hint:

Using similarity of triangles is much faster and less prone to quadratic equations than setting up algebraic variables like $AQ = x$ and using Pythagoras on $\Delta ACP$.

Answer 1

Step 1: Find PQ using Pythagoras, and locate C on OP.
In right triangle $OQP$ (right angle at $Q$, since $PQ$ is tangent), $PQ = \sqrt{OP^2 - OQ^2} = \sqrt{13^2-5^2} = \sqrt{144} = 12\text{ cm}$. Since $C$ is the point where tangent $AB$ touches the circle and lies on $OP$ with $OC$ a radius, $PC = OP - OC = 13 - 5 = 8\text{ cm}$.
Step 2: Set up an algebraic equation instead of using similar triangles.
Since $A$ lies on segment $PQ$ (between $P$ and $Q$) and $AC$, $AQ$ are both tangents from $A$, $AC = AQ$. Let $AC = AQ = t$, so $AP = PQ - AQ = 12 - t$.
Step 3: Apply Pythagoras in right triangle ACP.
Since $AB$ is tangent at $C$, $OC \perp AB$, so $\angle ACP = 90^\circ$. Then $AP^2 = AC^2 + CP^2$, giving $(12-t)^2 = t^2 + 8^2$. Expanding: $144 - 24t + t^2 = t^2 + 64$, so $144 - 24t = 64$, giving $24t = 80$ and $t = \frac{10}{3}\text{ cm}$.
Step 4: Compute AB and PA.
By the same tangent-length reasoning at $B$, $BC = AC = \frac{10}{3}\text{ cm}$, so $AB = AC + BC = \frac{20}{3}\text{ cm}$. Also $PA = 12 - t = 12 - \frac{10}{3} = \frac{26}{3}\text{ cm}$.
\[ \boxed{AB = \frac{20}{3}\text{ cm}, \ PA = \frac{26}{3}\text{ cm}} \]