Question

Let PA and PB be the tangent segments drawn from point P\((6,8)\) to the circle with the centre at origin O. The radius of circle for which the area of quadrilateral PAOB is maximum, is...

• A. \(5\)
• B. \(5\sqrt{2}\)
• C. \(\frac{5}{\sqrt{2}}\)
• D. \(\frac{5}{2}\)

Hint:

Area of PAOB equals r times the tangent length, so maximise r times sqrt(100 - r^2).

Answer 1

The Correct Option is B

Step 1: Use the angle at O
Let $\angle AOP = \phi$ with $\cos\phi = r/10$. The area of $PAOB$ is $2 \times \frac{1}{2}\, r \cdot OP \sin\phi = 10\, r \sin\phi$.

Step 2: Rewrite
With $r = 10\cos\phi$, area $= 100 \sin\phi\cos\phi = 50 \sin 2\phi$.

Step 3: Maximum
$\sin 2\phi$ is largest when $2\phi = 90^\circ$, so $\phi = 45^\circ$ and $r = 10\cos 45^\circ = 5\sqrt{2}$.

Step 4: Check
The maximum area is 50. Option (D) $r = \frac{5}{2}$ gives $\cos\phi = \frac{1}{4}$ and area $50\sin(2\phi) \approx 24.2$, which is less.

Final Answer:
The radius is 5 sqrt(2). This is option (B). \[ \boxed{\text{(B) }5\sqrt{2}} \]