Find the value of x in the given figure where TP (tangent at point T) = 15 cm,…

Concept — the tangent–secant (power of a point) relation. From a point outside a circle, suppose one line touches the circle at exactly one point (a tangent) and a second line through the same point cuts the circle at two points (a secant). Then the square of the tangent segment equals the product of the whole secant segment and its part lying outside the circle. Naming the external point P, the point of contact T, the nearer intersection A and the farther intersection B, the relation is PT2 = PA × PB. It is an equality between lengths only, so a figure with one unknown length turns into a single equation in that unknown.

Application — the given figure. Here P is the external point, PT = 15 cm is the tangent segment, PA = 9 cm is the part of the secant outside the circle, and PB = (2x + 3) cm is the whole secant from P through A to B.

1. Write the relation for this figure: PT2 = PA × PB.

2. Substitute the given lengths: 152 = 9 × (2x + 3).

3. Evaluate the square on the left: 225 = 9 × (2x + 3).

4. Divide both sides by 9: 25 = 2x + 3.

5. Subtract 3 from both sides: 22 = 2x.

6. Divide both sides by 2: x = 11.

Cross-check. Putting x = 11 back into the secant expression gives PB = 2(11) + 3 = 25 cm, so PA × PB = 9 × 25 = 225 cm2, while the tangent gives PT2 = 15 × 15 = 225 cm2. Both sides agree, so the length is consistent with the figure. The chord AB = PB − PA = 25 − 9 = 16 cm is positive, which matches A lying between P and B as drawn.

Common slips to avoid. Two errors change the equation itself: multiplying PA by the chord AB instead of by the whole secant PB, and forgetting to square the tangent length. Always pair the tangent squared with (external part) × (whole secant).