Find the equation of the circle which passes through the points (2, 3) and (4, 5)and the centre lies on the straight line y – 4x + 3 = 0.

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²----------- (A)

Putting (2,3) & (4,5) in the above equation,

(2 – h)² + (3 – k)² = r²

4 – 4h + h² + 9 + k² - 6k = r²

h² - 4h + k² - 6k + 13 = r² --------- (i)

(4 – h)² + (5 – k)² = r²

16 – 8h + h² + 25 + k² - 10k = r²

h² - 8h + k² - 10k + 41 = r² --------- (ii)

Equating both the equations (i) & (ii), as their RHS are equal,

h² - 4h + k² - 6k + 13 = h² - 8h + k² - 10k + 41

8h - 4h + 10k – 6k = 41 – 13

4h + 4k = 28

h + k = 7 ------- (iii)

As centre lies on the given line, so it satisfies the values too,

k – 4h + 3 = 0 ---------- (iv)

Solving equations {(iii) & (iii)} simultaneously,

h + k = 7

-4h + k = -3

Subtracting both the equations,

(+) (-) (+)

5h = 10

h = 2

2 + k = 7

k = 5

Putting h = 2 & k = 5 in equation (i),

h² - 4h + k² - 6k + 13 = r²

2² - 4(2) + 5² - 6(5) + 13 = r²

4 – 8 + 25 – 30 + 13 = r²

r² = 4

r = 2 units

Putting the values of h = 2, k = 5 & r = 2, respectively in equation (A),

(x – h)² + (y – k)² = r²

(x – 2)² + (y – 5)² = 2²

x² - 4x + 4 + y² - 10y + 25 = 4

x² - 4x + y² - 10y + 25 =0

Hence, the required equation is x² - 4x + y² - 10y + 25 =0.