Find the equation of the circle passing through the points A

Find the equation of the circle passing through the points A.(0,0) B. (12,4) C.(12, 18) The center5 of circle lies on each of perpendicular bisectors

Solution

General equation of circle is given by x^2+y^2+2gx+2fy+c=0
where g,f,c are constants.
Since given points pass through the circle then they must satisfy above equation of cirle
plug A(0,0) into x^2+y^2+2gx+2fy+c=0 gives
0^2+0^2+2g*0+2f*0+c=0
c=0 ...(i)

plug B(12,4) into x^2+y^2+2gx+2fy+c=0 gives
12^2+4^2+2g*12+2f*4+c=0
144+16+24g+8f+c=0
160+24g+8f+c=0
160+24g+8f+0=0 (using (i))
160+24g+8f=0
20+3g+f=0
f=-20-3g...(ii)

plug C(12,18) into x^2+y^2+2gx+2fy+c=0 gives
12^2+18^2+2g*12+2f*18+c=0
144+324+24g+36f+c=0
468+24g+36f+0=0 (using (i))
468+24g+36f=0
39+2g+3f=0...(iii)

Now we solve equation (ii) and (iii)
plug (ii) into (iii)
39+2g+3f=0
39+2g+3(-20-3g)=0
39+2g-60-9g=0
39+2g-60-9g=0
-7g-21=0
-7g=21
g=-3

now plug g=-3 into (ii)
f=-20-3g
f=-20-3(-3)
f=-20+9
f=-11

now plug values of f, g and c into x^2+y^2+2gx+2fy+c=0
x^2+y^2+2(-3)x+2(-11)y+0=0
x^2+y^2-6x-22y=0

Hence answer will be x^2+y^2-6x-22y=0