Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).

Let the coordinate of centre O be (x,y)

Then, OA = OB = OC {\(\because \) radius is equal}

Now,
OA = \( \sqrt{ (x-6)^2+(y+6)^2 } \)
OB = \( \sqrt{(x-3)^2 + (y+7)^2 } \)
OC = \( \sqrt{(x-3)^2 + (y-3)^2 } \)

Now, OA = OB
\( \Rightarrow \sqrt{(x-6)^2 + (y+6)^2 } = \sqrt{(x-3)^2 + (y+7)^2 } \)
\(\Rightarrow x^2 \ + \ 36 \ - \ 12x \ + \ y^2 \ + \ 36 \ - \ 12 \ = \ x^2 \ + \ 9 \ - \ 6x \ + \ y^2 \ + \ 9 \ - \ 6y \)
\( \Rightarrow 6x \ + 2y \ - \ 14 \ = \ 0 \) ..............(i)

Similarly, OB = OC
\( \Rightarrow \sqrt{(x-3)^2 + (y+7)^2 } = \sqrt{(x-3)^2 + (y-3)^2 } \)
\( \Rightarrow y^2 \ + \ 14y \ + \ 49 \ = \ y^2 \ - \ 6y \ + \ 9 \)
\(\Rightarrow y \ = \ -2 \)

Putting this value of y in eq. (i) , we get

\(\Rightarrow 6x \ + \ 2(-2) \ - \ 14 \ = \ 0 \)
\( \Rightarrow x \ = \ 3 \)

\(\therefore \) The centre of circle is at (3,-2)