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 a circle with center O
- \(\angle APB = 90^\circ\)
- \(AB = 3\sqrt{2}\) cm

To Find: Diameter of the circle

Step 1: Understand the geometry
Since PA and PB are tangents from point P and \(\angle APB = 90^\circ\), triangle APB is a right-angled triangle at P.
Also, OA and OB are radii and perpendicular to tangents, hence:
- \(OA \perp PA\) and \(OB \perp PB\)
- So, quadrilateral OAPB is a square-like structure with ∠APB = 90°

We now drop perpendiculars from center O to points A and B. Since triangle OAB is inside triangle APB and symmetric, triangle OAB is an isosceles right triangle, and ∠AOB = 90°.

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

Step 3: Find Diameter
\[ \text{Diameter} = 2r = 2 \times 3 = \boxed{6 \text{ cm}} \]

Final Answer:
The diameter of the circle is 6 cm.