Question

The centre of a circle of radius 13 units is the point (3, 6). P(7, 9) is a point inside the circle. APB is a chord of the circle such that AP = PB. Calculate the length of AB.

Answer 1

Given:

Centre of the circle, C = 3, 6

Radius of the circle, r = 13

Point P = 7, 9, inside the circle

APB is a chord such that AP = PB, which means P is the midpoint of chord AB.

Since P is inside the circle and AP = PB, P is the midpoint of chord AB.

The chord AB is bisected by the perpendicular line from the centre C to the chord AB at point P.

Use the distance formula to find the length of CP: 

`"Distance" = sqrt((x_2 − x_1)^2 + (y_2 − y_1)^2)`

The radius r = 13 is the hypotenuse in the right triangle CAP, where CP = 5 is one leg, and half the chord AP = PB = `(AB)/2` is the other leg.

Use Pythagoras theorem to find half the length of the chord AP: \[ AP = \sqrt{r^2 - CP^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\]

Therefore, the length of the chord \[AB = 2 \times AP = 2 \times 12 = 24 )\].

`AB = 2 xx AP`

= 2 × 12

= 24