Find the equation of the circle which passes

Solution:

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

(x – h)2 + (y – k)2 = r2………..1

Substituting (2, 3) & (4, 5) in the above equation, we get

(2 – h)2 + (3 – k)2 = r2

4 – 4h + h2 + 9 + k2 – 6k = r2

h2 – 4h + k2 – 6k + 13 = r2 …………..2

(4 – h)2 + (5 – k)2 = r2

16 – 8h + h2 + 25 + k2 – 10k = r2

h2 – 8h + k2 – 10k + 41 = r2 ………..3

Equating both the equations 2 & 3, as their RHS are equal, we get

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

On simplifying we get

8h – 4h + 10k – 6k = 41 – 13

4h + 4k = 28

h + k = 7 ………..4

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

k – 4h + 3 = 0…………5

Solving equations 3 and 4 simultaneously,

h + k = 7

-4h + k = -3

Subtracting both the equations, we get

5h = 10

h = 2

2 + k = 7

k = 5

Putting h = 2 & k = 5 in equation 2,

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 1,

(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.