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: Understand the figure
\(P = (6, 8)\) lies outside the circle with centre \(O\) at the origin. The tangents \(PA\) and \(PB\) are equal and each is perpendicular to the radius at its point of contact.

Step 2: Express the area
\(OP = \sqrt{36 + 64} = 10\). The tangent length is \(PA = \sqrt{OP^2 - r^2} = \sqrt{100 - r^2}\). The quadrilateral is two right triangles \(OAP\) and \(OBP\), so
\[ \text{Area} = 2 \times \frac{1}{2}\, r \sqrt{100 - r^2} = r\sqrt{100 - r^2} \]

Step 3: Maximise
The area is largest when its square \(r^2(100 - r^2)\) is largest. Let \(u = r^2\). Then \(u(100 - u)\) is maximum at \(u = 50\).

Step 4: Result
\(r^2 = 50\), so \(r = 5\sqrt{2}\). Then the area is \(5\sqrt{2} \cdot \sqrt{50} = 50\). Option (A) \(r = 5\) gives only \(5\sqrt{75} = 43.3\), and \(r = \frac{5}{\sqrt{2}}\) gives \(\frac{5}{\sqrt{2}}\sqrt{87.5} = 33.1\). So \(5\sqrt{2}\) is the maximum.

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