Question

Find the centre of the circle passing through $(6,-6),(3,-7)$ and $(3,3)$

Answer 1

Let the given points be $A(6,-6)$,$B(3,-7)$ and $C(3,3)$ and let the centre be $O(x,y)$

$OA=OB=OC$

So by distance formula we have,

Distance between two points = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

consider, $OA=OB$

$\Rightarrow {\left(OA\right)}^2={\left(OB\right)}^2$

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

$\Rightarrow x^2+36-12x+y^2+36+12y=x^2+9-6x+y^2+49+14y$

$\Rightarrow 3x+y=7$ ---(i)

Similarly, $OB=OC$

$\Rightarrow (OB{)}^2={\left(OC\right)}^2$

$\Rightarrow {\left(x-3\right)}^2+{\left(y+7\right)}^2={\left(x-3\right)}^2+{\left(y-3\right)}^2$

$\Rightarrow 14y+6y=18-58$

$\Rightarrow 20y=-40$

$\Rightarrow$$y=-2$

putting $y=-2$ in (i)

$3x+(-2)=7$

$\Rightarrow 3x-2=7$

$\Rightarrow$$x=3$

So, The centre of the circle is $(3,-2)$