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.
Show Hint
Remember: \(AQ=AC\) and \(BR=BC\). This is the standard trick for problems involving triangles circumscribing circles or tangents intersecting.
Step 1: Understanding the Concept:
1. Tangents from an external point to a circle are equal in length.
2. Tangent is perpendicular to the radius at the point of contact. Step 3: Detailed Explanation:
1. Finding PQ: In rt \(\triangle OQP\), \(OQ = 5\) cm (radius), \(OP = 13\) cm.
\[ QP^2 = OP^2 - OQ^2 = 13^2 - 5^2 = 169 - 25 = 144 \implies QP = 12 \text{ cm} \]
2. Since \(PQ\) and \(PR\) are tangents from \(P\), \(PQ = PR = 12\) cm.
3. Finding CP: \(OC\) is radius, so \(OC = 5\) cm.
\[ CP = OP - OC = 13 - 5 = 8 \text{ cm} \]
4. Since \(AB\) is a tangent at \(C\) on \(OP\), \(OP \perp AB\), so \(\triangle ACP\) is a right-angled triangle.
5. Let \(AC = x\). Since \(AC\) and \(AQ\) are tangents from point \(A\) to the circle:
\[ AQ = AC = x \]
6. Now, \(PA = PQ - AQ = 12 - x\).
7. In rt \(\triangle ACP\) (right angled at \(C\)):
\[ AP^2 = AC^2 + CP^2 \]
\[ (12 - x)^2 = x^2 + 8^2 \]
\[ 144 + x^2 - 24x = x^2 + 64 \]
\[ 24x = 144 - 64 = 80 \implies x = \frac{80}{24} = \frac{10}{3} \text{ cm} \]
8. Calculating lengths:
- \(PA = 12 - x = 12 - \frac{10}{3} = \frac{36 - 10}{3} = \frac{26}{3} \text{ cm} \).
- By symmetry, \(\triangle ACP \cong \triangle BCP\), so \(AC = BC = \frac{10}{3}\) cm.
- \(AB = AC + BC = \frac{10}{3} + \frac{10}{3} = \frac{20}{3} \text{ cm} \). Step 4: Final Answer:
The length of \(AB\) is \(\frac{20}{3}\) cm and the length of \(PA\) is \(\frac{26}{3}\) cm.
