A point P lies outside a circle with centre O such that OP = 6.5 cm. A secant PR intersects the circle at points Q and R (in that order from P), with PQ = 4.5 cm and QR = 3.5 cm. Find the radius of the circle in centimetres.

Introduction / Context:
This question tests the Power of a Point theorem for a point outside a circle. For a secant from an external point P that cuts the circle at Q and R, the product (PQ) * (PR) equals the power of the point. Also, when the circle center is O with OP known, the power equals OP^2 − r^2, where r is the radius.

Given Data / Assumptions:

• OP = 6.5 cm (distance from external point to center) • PQ = 4.5 cm (external segment of secant) • QR = 3.5 cm (segment inside the circle on the secant) • PR = PQ + QR

Concept / Approach:
Use Power of a Point: Power(P) = PQ * PR. Also for a circle with center O: Power(P) = OP^2 − r^2. Set them equal and solve for r. This directly converts the geometry into a simple algebraic equation.

Step-by-Step Solution:
1) Compute the whole secant length: PR = PQ + QR = 4.5 + 3.5 = 8.0 cm 2) Apply Power of a Point: Power(P) = PQ * PR = 4.5 * 8.0 = 36 3) Use center-distance relationship: Power(P) = OP^2 − r^2 4) Compute OP^2: OP^2 = 6.5^2 = 42.25 5) Set equal and solve: 42.25 − r^2 = 36 r^2 = 42.25 − 36 = 6.25 6) Take square root: r = √6.25 = 2.5 cm

Verification / Alternative check:
If r = 2.5, then OP^2 − r^2 = 42.25 − 6.25 = 36, which matches PQ*PR exactly. The data is consistent, so the radius is correct.

Why Other Options Are Wrong:
• 2, 3, 1.5, 4: squaring these radii would not satisfy OP^2 − r^2 = 36, so the power-of-point equality breaks.

Common Pitfalls:
• Using PQ*QR instead of PQ*PR. • Forgetting PR = PQ + QR (whole secant length). • Not squaring OP before subtracting r^2.

Final Answer:
2.5