Find the equation of the circle passing through the points (2, 3) and (–1,1) and
whose centre is on the line x – 3y – 11 = 0.

Let the equation of the circle be (x – h)² + (y – k)² = r²

Since the circle passes through (2, 3) and (-1, 1) we have

(2 – h)² + (3 – k)² = r² ……(i)

and (-1 – h)² + (1 – k)² = r² ……(ii)

Also since the centre lies on the line x – 3y – 11 = 0, we have

h – 3k = 11 …….(iii)

Simplifying the equation (i) and (ii) we have

(i) 4 – 4h + h² + 9 – 6k + k² = r²

13 – 4h + h² – 6k + k² = r²

(ii) 1 + 2h + h² + 1 – 2k + k² = r²

2 + 2h + h² – 2k + k² = r²

Eliminating square terms, we get

13 – 4h + h² – 6k + k² = 2 + 2h + h² – 2k + k²

or 13 – 2 – 4h – 2h + h² – h² – 6k + 2k + k² – k² = 0

or 11 – 6h – 4k = 0

or 6h + 4k = 11 ….(iv)

Solving (iii) and (iv) we get

6h – 18k = 66

6h + 4k = 11

- - -

– 22k = 55

or k = =

putting the value of k in (iii), we have

h – = 11

or h + = 11

or h = 11 – = =

Putting the value of (h, k) in (i), we get

+ = r²

or = r²

or = r²

or = r²

or = r²

or r² =

or r² =

Hence, the required equation of the circle is

=

or x² – 7x +

or x² – 7x + y² + 5y =

or x² + y² – 7x + 5y = 14

or x² + y² – 7x + 5y – 14 = 0.