For a complete understanding of the the solution given here, please go through the two files attached.
In the first pic all the required points have been shown. Let the center point be have the coordinates (h,k) as shown.
Since (h,k) is the center, it is equidistant from all the points. Let that distance be called "d".
Thus, using the distance formula, the distance, d from (h,k) to the point (2,2) is:
[tex] d=\sqrt{(2-h)^2+(2-k)^2} [/tex]
Squaring the above equation for convenience, we get:
[tex] d^2=(2-h)^2+(2-k)^2 [/tex]........................(Equation 1)
Likewise, for the point, (3,1), we get from the distance formula:
[tex] d^2=(3-h)^2+(1-k)^2 [/tex]........................(Equation 2)
Since these points are equidistant from (h,k), Equation 1 and Equation 2 will give us:
[tex] (2-h)^2+(2-k)^2=(3-h)^2+(1-k)^2 [/tex]
Expanding we get:
[tex] 4+h^2-4h+4+k^2-4k=9+h^2-6h+1+k^2-2k [/tex]
Cancelling the common terms we get:
[tex] 8-4h-4k=10-6h-2k [/tex]
Rearranging, we get: [tex] -4h+gh-4k+2k=10-8 [/tex]
[tex] 2h-2k=2 [/tex]
[tex] h-k=1 [/tex].......................(Equation 3)
Likewise, if we take the points (3,1) and (2,0) which are equidistant from (h,k), we will get:
[tex] (3-h)^2+(1-k)^2=(2-h)^2+(0-k)^2 [/tex]
[tex] 9+h^2=6h+1+k^2-2k=4+h^2-4h+k^2 [/tex]
Rearrangement yields:
[tex] -2h-2k=-6 [/tex]
Or, [tex] h+k=3 [/tex].................................(Equation 4)
Adding (Equation 3) and (Equation 4) we will get:
[tex] 2h=4 [/tex]
Thus, [tex] h=2 [/tex]
Plugging in this value of h in (Equation 3) we get:
[tex] 2-k=1 [/tex]
or [tex] k=2-1=1 [/tex]
Thus, the required coordinates are: [tex] (h,k)=(2,1) [/tex]
The complete information can be got from the second file attached. As we can see the complete diagram is that of a circle.
