Question

Find the equation of circle which passes through the points (3 , 0) , ( 1 , -6) and (4 , -1)?

Answer 1

First Method :-
$x^2+y^2+2gx+2fr+c=0.$
It passes through the points (3, 0), (1, -6) and (4, -1).
$\therefore$ 6g + 0f + 9 + c = 0 .(1)
2g - 12f + 37 + c = 0 .(2)
8g - 2f + 17 + c = 0. ..(3)
Subtracting (2) from both (1) and (3) thereby eliminating c,
4g + 12f 28 = 0
or g + 3f 7 = 0 ..(4)
6g + 10f 20 = 0
or 3g + 5f 10 = 0 .(5)
Solving (4) and (5), we get
f = 11 / 4, g = - 5 / 4
and hence from (1), c = - 3 / 2.
Putting for g , f and c the required circle is
$2x^2+2y^2-5x+11y$ 3 = 0.
Second Method :- When three points A, B, C are given , we write the equation of a circle on AB as diameter, say S = 0 and also the equation of AB, say u = 0.Then any circle through A ,B will be of the form S + $\lambda$u = 0. Then $\lambda$ is determined by the condition that it passes through C. Here A, B, C are(3, 0), (1, - 6) and (4, -1) respectively. The circle on AB.as diameter is
S = (x - 3) (x - 1) + (y - 0) (y + 6) = 0
or S = $x^2$ + $y^2$ - 4x + 6y + 3 = 0
And equation of the line AB is
y 0 = $\frac{-6-0}{1-3}$,(x - 3)
or u = 3x y 9 = 0
$\therefore$ Any circle through A, B is S + $\lambda$u = 0
$x^2$ +$y^2$ - 4x + 6y + 3 + $\lambda$ (3x y 9) = 0
It passes through (4, -1) so that
16 + 1 16 6 + 3 + $\lambda$ (12 + 1 - 9) = 0
$\therefore \lambda =\frac{2}{4}=\frac{1}{2}$
Hence the required equation of the circle is
$x^2+y^2-4x+6y+3+\frac{1}{2}$ (3x - y - 9) = 0
or $2x^2$ + $2y^2$ - 5x + 11y 3 = 0