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

Given:

The centre of a circle passes through $(6, -6), (3, -7)$ and $(3, 3)$.

To do:

We have to find the centre of the given circle.

Solution:

Let \( \mathrm{O} \) is the centre of the circle and \( \mathrm{A}(6,-6), \mathrm{B} (3,-7) \) and \( \mathrm{C}(3,3) \) are the points on the circle.

Let the coordinates of \( \mathrm{O} \) are \( (x, y) \).

This implies,

\( \mathrm{OA}=\mathrm{OB}=\mathrm{OC} \)         (Radii of the circle)

\( \mathrm{OA}^{2}=\mathrm{OB}^{2}=\mathrm{OC}^{2} \) 

On squaring, we get,

\( \mathrm{OA}^{2}=(x-6)^{2}+(y+6)^{2} \)

\( =x^{2}-12 x+36+y^{2}+12 y+36 \)

\( =x^{2}+y^{2}-12 x+12 y+72 \)

\( \mathrm{OB}^{2}=(x-3)^{2}+(y+7)^{2} \)

\( =x^{2}+9-6 x+y^{2}+49+14 y \)

\( =x^{2}+y^{2}-6 x+14 y+58 \)

\( \mathrm{OC}^{2}=(x-3)^{2}+(y-3)^{2} \)

\( =x^{2}+9-6 x+y^{2}-6 y+9 \)

\( =x^{2}+y^{2}-6 x-6 y+18 \)

\( \mathrm{OA}^{2}=\mathrm{OB}^{2} \)

\( \Rightarrow x^{2}+y^{2}-12 x+12 y+72=x^{2}+y^{2}-6 x+14 y+58 \)

\( \Rightarrow -12x+6 x+12 y-14 y=58-72 \)

\( \Rightarrow -6 x-2 y=-14 \)

\( \Rightarrow -2(3 x+y)=-2(7) \)

\( \Rightarrow 3 x+y=7 \).........(i)

\( \mathrm{OB}^{2}=\mathrm{OC}^{2} \)

\( \Rightarrow x^{2}+y^{2}-6 x+14 y+58=x^{2}+y^{2}-6 x-6 y+18 \)

\( \Rightarrow 14 y+6 y=18-58 \)

\( \Rightarrow 20 y=-40 \)

\( \Rightarrow y=\frac{-40}{20}=-2 \)

Substituting the value of \( y \) in (i), we get,

\( \Rightarrow 3 x-2=7 \)

\( \Rightarrow 3 x=7+2=9 \)

\( \Rightarrow x=\frac{9}{3}=3 \) 

Therefore, the centre of the given circle is $(3, -2)$.