Find the centre of the circle passing through

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

The centre of a circle is at equal distance from all the points on its circumference.

Here it is given that the circle passes through the points A(6,−6), B(3,−7) and C(3,3).

Let the centre of the circle be represented by the point O(x, y).

So we have 

$A O=\sqrt{(6-x)^{2}+(-6-y)^{2}}$

$B O=\sqrt{(3-x)^{2}+(-7-y)^{2}}$

$C O=\sqrt{(3-x)^{2}+(3-y)^{2}}$

Equating the first pair of these equations we have,

$A O=B O$

$\sqrt{(6-x)^{2}+(-6-y)^{2}}=\sqrt{(3-x)^{2}+(-7-y)^{2}}$

Squaring on both sides of the equation we have,

$(6-x)^{2}+(-6-y)^{2}=(3-x)^{2}+(-7-y)^{2}$

$36+x^{2}-12 x+36+y^{2}+12 y=9+x^{2}-6 x+49+y^{2}+14 y$

$6 x+2 y=14$

$3 x+y=7$

Equating another pair of the equations we have,

$A O=C O$

$\sqrt{(6-x)^{2}+(-6-y)^{2}}=\sqrt{(3-x)^{2}+(3-y)^{2}}$

Squaring on both sides of the equation we have,

$(6-x)^{2}+(-6-y)^{2}=(3-x)^{2}+(3-y)^{2}$

$36+x^{2}-12 x+36+y^{2}+12 y=9+x^{2}-6 x+9+y^{2}-6 y$

$6 x-18 y=54$

$x-3 y=9$

Now we have two equations for ‘x’ and ‘y’, which are

$3 x+y=7$

$x-3 y=9$

From the second equation we have. Substituting this value of ‘y’ in the first equation we have,

$x-3(-3 x+7)=9$

$x+9 x-21=9$

$10 x=30$

$x=3$

Therefore the value of ‘y’ is,

$y=-3 x+7$

$=-3(3)+7$

$y=-2$

Hence the co-ordinates of the centre of the circle are $(3,-2)$.