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: Understanding the Question:
We are given a circle with center $O$ and radius $5\text{ cm}$.
$PQ$ and $PR$ are tangents from an external point $P$, so $OQ \perp PQ$.
Another tangent $AB$ touches the circle at $C$, which lies on the line segment $OP$.
The total distance is $OP = 13\text{ cm}$. We need to find the lengths of $AB$ and $PA$.

Step 2: Key Formula or Approach:
- In right-angled triangle $\Delta OQP$, use Pythagoras theorem: $PQ = \sqrt{OP^2 - OQ^2}$.
- Tangents from an external point to a circle are equal: $AQ = AC$.
- Use similar triangles $\Delta ACP \sim \Delta OQP$ to find the unknown segments.

Step 3: Detailed Explanation:

• Find the length of the tangent $PQ$ using Pythagoras' theorem in right triangle $\Delta OQP$:
\[ PQ = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm} \]

• Find the distance $PC$:
Since $C$ is a point of tangency on $OP$, the segment $OC$ is a radius:
\[ OC = 5\text{ cm} \]
Therefore:
\[ PC = OP - OC = 13 - 5 = 8\text{ cm} \]

• Prove similarity between right-angled triangles $\Delta ACP$ and $\Delta OQP$:
- $\angle ACP = \angle OQP = 90^\circ$
- $\angle APC = \angle OPQ$ (Common angle)
By AA similarity:
\[ \Delta ACP \sim \Delta OQP \]

• Set up ratios of corresponding sides:
\[ \frac{AC}{OQ} = \frac{PC}{PQ} \]
Substitute the known values ($OQ = 5$, $PC = 8$, $PQ = 12$):
\[ \frac{AC}{5} = \frac{8}{12} = \frac{2}{3} \]
\[ AC = \frac{10}{3}\text{ cm} \]

• Find the length $AB$:
Due to symmetry of the tangents from $A$ and $B$:
\[ BC = AC = \frac{10}{3}\text{ cm} \]
Therefore:
\[ AB = AC + BC = \frac{10}{3} + \frac{10}{3} = \frac{20}{3}\text{ cm} \]

• Find the length $PA$:
Since $AQ$ and $AC$ are tangents from external point $A$:
\[ AQ = AC = \frac{10}{3}\text{ cm} \]
Therefore:
\[ PA = PQ - AQ = 12 - \frac{10}{3} = \frac{36 - 10}{3} = \frac{26}{3}\text{ cm} \]


Step 4: Final Answer:
The length of $AB$ is $\frac{20}{3}\text{ cm}$ (or $6.67\text{ cm}$) and the length of $PA$ is $\frac{26}{3}\text{ cm}$ (or $8.67\text{ cm}$).