Two chords AB, CD of a circle intersect internally at a point P. If

(i) AP = 6 cm, PB = 4 cm and PD = 3 cm, find PC.

(ii) AB = 12 cm, AP = 2 cm, PC = 5 cm, find PD.

(iii) AP = 5 cm, PB = 6 cm and CD = 13 cm, find CP.

We know that when two chords of a circle intersect internally or externally, then the products of the lengths of segments are equal.

Given, chords AB and CD of a circle intersect internally at a point P. So,

PA.PB = PC.PD

(i) Given, AP = 6 cm, PB = 4 cm and PD = 3 cm.

We know that,

PA.PB = PC.PD

⇒ 6 × 4 = PC × 3
⇒ 24 = 3PC
⇒ PC = $\dfrac{24}{3}$ = 8 cm.

Hence, the length of PC = 8 cm.

(ii) Given, AB = 12 cm, AP = 2 cm, PC = 5 cm

PB = AB - AP = 12 - 2 = 10 cm.

We know that,

PA.PB = PC.PD

⇒ 2 × 10 = 5 × PD
⇒ 20 = 5PD
⇒ PD = $\dfrac{20}{5}$ = 4 cm.

Hence, the length of PD = 4 cm.

(iii) Given, AP = 5 cm, PB = 6 cm and CD = 13 cm

Let PC = x, so PD = 13 - x

We know that,

PA.PB = PC.PD

⇒ 5 × 6 = x(13 - x)
⇒ 30 = 13x - x²
⇒ x² - 13x + 30 = 0
⇒ x² - 10x - 3x + 30 = 0
⇒ x(x - 10) - 3(x - 10) = 0
⇒ (x - 3)(x - 10) = 0
⇒ x - 3 = 0 or x - 10 = 0
⇒ x = 3 or x = 10.

Hence, PC = 3 cm or 10 cm.