How to find Equation of a Circle if co-ordinates of its three points are given
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.
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