Question

In the adjoining figure, PA and PB are tangents to a circle with centre O such that $\angle P = 90^\circ$. If $AB = 3\sqrt{2}$ cm, then the diameter of the circle is

• A. $3\sqrt{2}$ cm
• B. 6 cm
• C. 3 cm
• D. $6\sqrt{2}$ cm

Hint:

When two tangents from a point form a right angle, the line joining points of contact becomes the diagonal of a square whose side is the radius.

Answer 1

The Correct Option is B

Given:
- PA and PB are tangents from point P to circle O
- \(\angle APB = 90^\circ\)
- \(AB = 3\sqrt{2}\) cm

To Find: Diameter of the circle

Solution:
Since PA and PB are tangents and \(\angle APB = 90^\circ\), triangle APB is right-angled at P. OA and OB are radii, thus:
- \(OA \perp PA\) and \(OB \perp PB\)
- Quadrilateral OAPB resembles a square with ∠APB = 90°

Triangle OAB is an isosceles right triangle, and ∠AOB = 90°.

In triangle AOB:
- AB is the hypotenuse = \(3\sqrt{2}\) cm
- Let radius \(r = OA = OB\)
Using Pythagoras Theorem:
\[AB^2 = OA^2 + OB^2 = r^2 + r^2 = 2r^2\Rightarrow (3\sqrt{2})^2 = 2r^2\Rightarrow 9 \cdot 2 = 2r^2\Rightarrow 18 = 2r^2\Rightarrow r^2 = 9\Rightarrow r = 3 \text{ cm}\]

The diameter is:
\[\text{Diameter} = 2r = 2 \times 3 = \boxed{6 \text{ cm}}\]

Answer: The diameter is 6 cm.