In the given figure, PA is the tangent to the circle with centre O such that OA = 10 cm, AB = 8 cm and AB \(\perp\) OP. Find the length of PB.
Show Hint
An important property of right-angled triangles to remember:
The square of the altitude to the hypotenuse equals the product of the two segments of the hypotenuse:
\[ AB^2 = OB \times PB \]
Substituting the values immediately gives \( 8^2 = 6 \times PB \implies 64 = 6 \times PB \implies PB = \frac{32}{3}\text{ cm} \).
This direct geometric relation avoids setting up full triangle similarity statements.
Step 1: Set up the right triangle and find OB.
We are given a circle with centre $O$ and radius $OA = 10$ cm. $PA$ is a tangent at $A$, so $OA \perp PA$, giving $\angle OAP = 90^{\circ}$.
$AB$ is drawn perpendicular to $OP$, with $AB = 8$ cm, so $\angle OBA = \angle ABP = 90^{\circ}$.
In right triangle $OBA$ (right angled at $B$), apply the Pythagoras theorem:
\[ OA^2 = OB^2 + AB^2 \]
\[ 10^2 = OB^2 + 8^2 \]
\[ 100 = OB^2 + 64 \]
\[ OB^2 = 36 \implies OB = 6\text{ cm} \]
Step 2: Recognize AB as the altitude to the hypotenuse of triangle OAP.
Because $PA$ is a tangent, $\angle OAP = 90^{\circ}$, so triangle $OAP$ is right angled at $A$, with $OP$ as its hypotenuse.
Since $AB \perp OP$, the segment $AB$ is exactly the altitude drawn from the right angle vertex $A$ to the hypotenuse $OP$.
There is a known geometric mean (mean proportional) rule for this exact setup: when the altitude is drawn from the right angle to the hypotenuse, splitting the hypotenuse into two parts, the square of the altitude equals the product of those two parts.
Here the hypotenuse $OP$ is split by $B$ into $OB$ and $BP$, so this rule gives directly:
\[ AB^2 = OB \times PB \]
Step 3: Substitute the known values into the geometric mean relation.
We know $AB = 8$ cm and $OB = 6$ cm from Step 1. Substitute into $AB^2 = OB \times PB$:
\[ 8^2 = 6 \times PB \]
\[ 64 = 6 \times PB \]
\[ PB = \frac{64}{6} = \frac{32}{3}\text{ cm} \]
Final Answer:
Using the altitude-on-hypotenuse geometric mean relation instead of setting up a full similar triangle ratio, the length of PB comes out to the same value.
\[ \boxed{PB = \frac{32}{3}\text{ cm} \approx 10.67\text{ cm}} \]
