###### Advertisements

###### Advertisements

In the following figure, a tangent segment PA touching a circle in A and a secant PBC is shown. If AP = 15, BP = 10, find BC.

###### Advertisements

#### Solution

PA is a tangent segment and PBC is the secant.

`therefore PB xx PC=PA^2` ........(Relationship to Tangent-Secant Theorem)

`therefore 10xxPC=15^2`

`PC=225/10=22.5`

`Now , PB+BC=PC`

`therefore 10+BC=22.5`

`therefore BC=22.5-10= 12.5`

#### APPEARS IN

#### RELATED QUESTIONS

In the following figure, ray PA is a tangent to the circle at A and PBC is a secant. If AP = 15, BP = 10, then find BC.

In the given figure, P is the point of contact.

(1) If m(arc PR) = 140°, ∠ POR = 36°, find m(arc PQ)

(2) If OP = 7.2, OQ = 3.2, find OR and QR

(3) If OP = 7.2, OR = 16.2, find QR.

In the figure given above, ‘O’ is the centre of the circle, seg PS is a tangent segment and S is the point of contact. Line PR is a secant.

If PQ = 3.6, QR = 6.4, find PS.

**Solution:**

PS^{2} = PQ × `square` ......(tangent secant segments theorem)

= PQ × (PQ + `square`)

= 3.6 × (3.6 + 6.4)

= 3.6 × `square`

= 36

∴ PS = `square` .....(by taking square roots)