Find the equation of the circle passing through the points (4,1) and (6,5) and
whose centre is on the line 4x + y = 16.

Since the circle passes through (4, 1) and (6, 5), we have

(4 – h)² + (1 – k)² = r²

and (6 – h)² + (5 – k)² = r²

Also, since the centre lies on the line 4x + y = 16, we have

4h + k = 16

Simplifying the equation (i), we get

16 – 8h + h² + 1 – 2k + k² = r²

or 61 – 12h + h² – 10k + k² = r²

Now,

17 – 61 – 8h + 12h + h² - 2k + 10k + k² – k² = 0

(eliminating square terms)

or -44 + 4h + 8k = 0

or 4h + 8k = 44

solving equation (iii) and (iv) we get

4h + k = 16

4h + 8k = 44

- - -

-7k = -28 ⇒ k = 4

Substituting k = 4 in equation (iii) we get

4h + 4 = 16

or 9; 9; 9; 4h = 12

or h = 3

Substituting the value of h = 3, k = 4 in equation

(i) we get

(4 – 3)² + (1 – 4)² = r²

or 1 + 9 = r²

or 10 = r²

Hence, the required equation of the circle is

(4 – 3)² + (y – 4)² = 10

or x² – 6x + 9 +y² – 8y + 16 = 10

or x² – y² – 6x – 8y + 15 = 0.