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 PQ using Pythagoras, and PC using the radius. \[ PQ = \sqrt{OP^2-OQ^2} = \sqrt{13^2-5^2} = \sqrt{144}=12\text{ cm} \] Since $C$ lies on $OP$ with $OC=5\text{ cm}$ (radius), $PC = OP-OC = 13-5=8\text{ cm}$. Step 2: Use the tangent of the shared angle instead of writing a full similarity proportion.
In right triangle $OQP$ (right-angled at $Q$), $\tan(\angle OPQ) = \frac{OQ}{PQ} = \frac{5}{12}$. Since $\angle ACP = 90^\circ$ ($AB \perp OP$ at the point of tangency $C$) and $\angle APC = \angle OPQ$ is the same angle, triangle $ACP$ has the same tangent ratio: $\tan(\angle APC) = \frac{AC}{PC}$. Step 3: Solve for AC, then AB. \[ \frac{AC}{8} = \frac{5}{12} \] \[ AC = \frac{40}{12} = \frac{10}{3}\text{ cm} \] By symmetry about $OP$, $BC = AC = \frac{10}{3}\text{ cm}$, so \[ AB = AC+BC = \frac{20}{3}\text{ cm} \] Step 4: Find PA using the equal-tangent property.
Since $AQ$ and $AC$ are tangents from the same external point $A$, $AQ = AC = \frac{10}{3}\text{ cm}$, so \[ PA = PQ - AQ = 12 - \frac{10}{3} = \frac{26}{3}\text{ cm} \]
\[ \boxed{AB = \frac{20}{3}\text{ cm}, \ PA = \frac{26}{3}\text{ cm}} \]
