find the centre of the circle …

Let A → (6, –6), B → (3, –7) and C → (3, 3).
Let the centre of the circle be I(x, y)
Then, IA = IB = IC [By definition of a circle]
{tex}\Rightarrow{/tex} IA² = IB² = IC²
{tex}\Rightarrow{/tex} (x - 6)² + (y + 6)² = (x - 3)² + (y + 7)² = (x - 3)² + (y - 3)²
Taking first two, we get
(x - 6)² + (y + 6)² = (x - 3)² + (y + 7)²
{tex}\Rightarrow{/tex} x² - 12x + 36 + y² + 12y + 36  = x² - 6x + 9 + y² + 14y + 49
{tex}\Rightarrow{/tex} 6x + 2y = 14
{tex}\Rightarrow{/tex} 3x + y = 7 ......(1) ....[Dividing throughout by 2]
Taking last two, we get
(x - 3)² + (y + 7)² = (x - 3)² + (y - 3)²
{tex}\Rightarrow{/tex} (y + 7)² = (y - 3)²
{tex}\Rightarrow{/tex} (y + 7) = {tex}\pm{/tex}(y-3)
taking +e sign, we get
y + 7 = y - 3
{tex}\Rightarrow{/tex} 7 = -3
which is impossible
Taking -ve sign, we get
y + 7 = -(y - 3)
{tex}\Rightarrow{/tex} y + 7 = -y + 3
{tex}\Rightarrow{/tex} 2y = -4
{tex}\Rightarrow y = \frac{{ - 4}}{2} = - 2{/tex}
Putting y = -2 in equation (1), we get
{tex}\Rightarrow{/tex} 3x - 2 = 7
{tex}\Rightarrow{/tex} 3x = 9
{tex}\Rightarrow{/tex} x = 3
Thus, I {tex}\rightarrow{/tex} (3, -2)
Hence, the centre of the circle is (3, -2).