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.
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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 $\Delta ACP$.
Step 1: Find tangent length $PQ$ using Pythagoras in right triangle $OQP$. \[ PQ = \sqrt{OP^2-OQ^2} = \sqrt{13^2-5^2} = \sqrt{144} = 12\text{ cm} \] Step 2: Use the tangent-length relation directly, treating $A$ as an external point to the circle with two tangents $AQ$ and $AC$. Since $C$ lies on $OP$ with $OC=5\text{ cm}$ (radius), $PC = OP-OC = 8\text{ cm}$. By similar right triangles $ACP$ and $OQP$ (both share angle $P$ and have a right angle at $C$ and $Q$ respectively), the sides scale by the same factor $\frac{PC}{PQ}=\frac{8}{12}=\frac23$. Step 3: Apply this scale factor to find $AC$, and hence $AQ$ (equal tangents from $A$). \[ AC = \frac23 \times OQ = \frac23 \times 5 = \frac{10}{3}\text{ cm} = AQ \] By the same logic on the other side, $BC=\frac{10}{3}\text{ cm}$, so $AB = AC+BC = \frac{20}{3}\text{ cm}$. Step 4: Find $PA$ by subtracting the equal tangent $AQ$ from $PQ$, and conclude. \[ PA = PQ - AQ = 12 - \frac{10}{3} = \frac{26}{3}\text{ cm} \] \[ \boxed{AB = \frac{20}{3}\text{ cm},\ PA = \frac{26}{3}\text{ cm}} \]
