How to find equation of the circle passing through (9,1), (-3,-1) and (4,5)
The equation of a circle is more specific: (x-h)^2+(x-k)^2=a^2, where (h,k) is the centre and a the radius.
Plug in the points:
This can be written (9-h)^2+(1-k)^2=(3+h)^2+(1+k)^2=(4-h)^2+(5-k)^2=a^2
Leaving a out of it for the moment we can use pairs of equations, using difference of squares:
12(6-2h)+2(-2k)=0, 36-12h-2k=0, 18-6h-k=0, so k=18-6h.
7(-1-2h)+6(-4-2k)=0, -7-14h-24-12k=0, -31-14h-12k=0 or 31+14h+12k=0, 31+14h+12(18-6h)=0, 31+14h+216-72h=0.
247-58h=0 so h=247/58
This looks suspiciously ungainly. So rather than continuing, I'm going to call the three points: (Q,R), (S,T), (U,V) to provide a general answer to all questions of this sort.
[(equation 1)=(equation 2)=(equation 3)=a^2]
Take the equations in pairs and temporarily ignore a^2.
Equations 1 and 2:
2k(R-T)=Q^2+R^2-(S^2+T^2)-2h(Q-S), so k=(Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T))
Equations 2 and 3:
S^2-U^2-2h(S-U)+T^2-V^2-2k(T-V)=0, so k=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V))
At this point, we can substitute for k and end up with an equation involving the unknown h only:
Once h is found we can calculate k, and then we can substitute the values for h and k in any equation to find a^2 which is equal to each of the three equations.
When we use Q=9, R=1, S=-3, T=-1, U=4, V=5 or -5, we appear to get very ungainly solutions. One way to find which values need to be changed may be to plot the values and work out where the circle should fit and where its centre has less complex values. If V=6 or -6, there is a simple solution: (x-3)^2+y^2=37 (centre at (3,0), a^2=37=6^2+1^2. (-3-3)^2=(9-3)^2=6^2; (-6)^2=6^2; (2-3)^2=(4-3)^2=1; (-1)^2=1^2 shows the combination of points that would lie on the circle: (9,1), (9,-1), (-3,1), (-3,-1), (4,6), (4,-6), (2,6), (2,-6) and this includes points A and B, but not C. Note that the equation k=18-6h is valid for h=3, k=0. Read More: ...