Find the centre of the circle passing through (5, −8),

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(5,−8), B(2,−9) and C(2,1).

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

So we have 

$A O=\sqrt{(5-x)^{2}+(-8-y)^{2}}$

$B O=\sqrt{(2-x)^{2}+(-9-y)^{2}}$

$C O=\sqrt{(2-x)^{2}+(1-y)^{2}}$

Equating the first pair of these equations we have,

$A O=B O$

$\sqrt{(5-x)^{2}+(-8-y)^{2}}=\sqrt{(2-x)^{2}+(-9-y)^{2}}$

Squaring on both sides of the equation we have,

$(5-x)^{2}+(-8-y)^{2}=(2-x)^{2}+(-9-y)^{2}$

$25+x^{2}-10 x+64+y^{2}+16 y=4+x^{2}-4 x+81+y^{2}+18 y$

$6 x+2 y=4$

$3 x+y=2$

Equating another pair of the equations we have,

$A O=C O$

$\sqrt{(5-x)^{2}+(-8-y)^{2}}=\sqrt{(2-x)^{2}+(1-y)^{2}}$

Squaring on both sides of the equation we have,

$(5-x)^{2}+(-8-y)^{2}=(2-x)^{2}+(1-y)^{2}$

$25+x^{2}-10 x+64+y^{2}+16 y=4+x^{2}-4 x+1+y^{2}-2 y$

$6 x-18 y=84$

$x-3 y=14$

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

$3 x+y=2$

$x-3 y=14$

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

$x-3(-3 x+2)=14$

$x+9 x-6=14$

$10 x=20$

$x=2$

Therefore the value of ‘y’ is,

$y=-3 x+2$

$=-3(2)+2$

$y=-4$

Hence the co-ordinates of the centre of the circle are $(2,-4)$.