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Find the centre of a circle passing through the points $(6, -6), (3, -7)$ 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 $\mathrm{O}$ be 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 $\mathrm{O}$ be $(x, y)$.
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,
$3 x-2=7$
$\Rightarrow 3 x=7+2=9$
$\Rightarrow x=\frac{9}{3}=3$
Therefore, the centre of the given circle is $(3, -2)$.