Find the centre of a circle …

Let the given points are A(6,-6), B(3,-7), C(3,3)
And P(x,y) be the centre of the circle. So, AP = BP = CP (radii of the circle)

Taking AP = BP and Squaring both sides, we get,
{tex}\Rightarrow A P ^ { 2 } = B P ^ { 2 }{/tex}
{tex}\Rightarrow ( x - 6 ) ^ { 2 } + ( y + 6 ) ^ { 2 } = ( x - 3 ) ^ { 2 } + ( y + 7 ) ^ { 2 }{/tex}(by using distance formula)
{tex}\Rightarrow x ^ { 2 } - 12 x + 36 + y ^ { 2 } + 12 y + 36{/tex} ={tex}x ^ { 2 } - 6 x + 9 + y ^ { 2 } + 14 y + 49{/tex}
{tex}\Rightarrow - 12 x + 6 x + 12 y - 14 y + 72 - 58 = 0{/tex}
{tex}\Rightarrow - 6 x - 2 y + 14 = 0{/tex}
{tex}\Rightarrow 3 x + y - 7 = 0{/tex} ……….(i)

Again, taking BP = CP and squaring both sides, we get,
{tex}\Rightarrow B P ^ { 2 } = C P ^ { 2 }{/tex}
{tex}\Rightarrow ( x - 3 ) ^ { 2 } + ( y + 7 ) ^ { 2 } = ( x - 3 ) ^ { 2 } + ( y - 3 ) ^ { 2 }{/tex}
{tex}\Rightarrow x ^ { 2 } - 6 x + 9 + y ^ { 2 } + 14 y + 49{/tex} ={tex}x ^ { 2 } - 6 x + 9 + y ^ { 2 } - 6 y + 9{/tex}
{tex}\Rightarrow - 6 x + 6 x + 14 y + 6 y + 58 - 18 = 0{/tex}
{tex}\Rightarrow 20 y + 40 = 0{/tex}
{tex}\Rightarrow y = - 2{/tex}

Putting the value of y in eq. (i),
{tex}3x + y - 7 = 0{/tex}
{tex}\Rightarrow 3 x = 9{/tex}
{tex}\Rightarrow x = 3{/tex}
Hence the coordinates of P i.e. centre of circle are (3,-2).