# 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 $(3, -2)$.

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Updated on: 10-Oct-2022

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