How to find Equation of a Circle by its three points

QUESTION: What is the equation of a circle passing through the points (0,0), (-5,4) and (6,10)?

SOLUTION:

As we know, the general equation of a circle is

x² + y² + 2gx + 2fy + c = 0 ——————(1)

where (-g, -f) is the center of a circle and √g² + f² – c is the radius.

According to the given problem, the circle passes through the points (0,0), (-5,4) and (6,10). Therefore

For point (0, 0), put x= 0 and y= 0 in equation (1)

(0)² + (0)² + 2g(0) + 2f(0) + c = 0

which implies

c = 0 —————————————(2)

For point (-5, 4), put x= -5 and y= 4 in equation (1)

(-5)² + (4)² + 2g(-5) + 2f(4) + c = 0

which implies

25 + 16 -10g + 8f + c = 0

41 – 10g + 8f + c = 0

From equation (2), c = 0, so above equation reduces to

41 – 10g + 8f = 0

-10g + 8f = -41 —————————————(3)

For point (6, 10), put x= 6 and y= 10 in equation (1)

(6)² + (10)² + 2g(6) + 2f(10) + c = 0

which implies

36 + 100 + 12g + 20f + c = 0

136 + 12g + 20f + c = 0

From equation (2), c = 0 so above equation reduces to

136 + 12g + 20f = 0

12g + 20f = -136

3g + 5f = -34—————————————(4)

Solve equations (3) and (4) by substitution to find the value of g and f

From equation (3)

g = (41 / 10) + (4/5)f

Substitute value of g in equation (4), we get

f = $\frac{-463}{74}$ ———————————(5)

Now put value of f in equation (4) to get value of g, we get

g = $\frac{-67}{74}$ ———————————(6)

Put value of g, f, c from equations (6), (5) and (2) in equation (1)

x² + y² + 2gx + 2fy + c = 0 ——————(1)

x² + y² + 2( $\frac{-67}{74}$ )x + 2( $\frac{-463}{74}$ )y = 0

x² + y² – ( $\frac{67}{37}$ )x – ( $\frac{463}{37}$ )y = 0

37x² + 37y² – 67x – 463y = 0

This is the required equation of circle.