Question

C(10, 4) is the centre of the circle with radius 17 units. CM ⊥ chord AB and M ≡ (1, −8). Calculate the lengths of AM and AB.

Answer 1

Centre of the circle C(10, 4)

Radius (r = 17) units

CM ⊥ AB, i.e., CM is perpendicular to chord AB.

Point M = (1, −8)

\[ CM = \sqrt{(10-1)^2 + (4-(-8))^2} \]

= \[\sqrt{9^2 + 12^2} \]

= \[\sqrt{81 + 144}\]

= \[\sqrt{225} \]

= 15 units

Use the right triangle formed by C, M, and A or B to find half chord length AM

Let AM = BM = x since CM perpendicular bisects chord AB.

Using Pythagoras theorem in triangle CMA:

CA² = CM² + AM² 

We know radius CA = 17.

Substitute:

17² = 15² + x² 

289 = 225 + x²

x² = 289 − 225 = 64 

x = 8

Now AM = 8 units.

Length of chord AB = 2 × AM = 2 × 8 = 16 units.