Question

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

• A. $(1,-2)$
• B. $(3,-2)$
• C. $(0,-2)$
• D. None of these

Answer 1

The correct option is B $(3,-2)$

Let the circle pass through points $A(6,-6),B(3,-7)$ and $C(3,3)$

Let $O(x,y)$ be the centre of the circle.

Since the centre is equidistant from all the points lying on the circle.

Distance $AO=$Distance $CO$

Using distance formula,

$\sqrt{{(x-6)}^2+{(y+6)}^2}=\sqrt{{(x-3)}^2+{(y-3)}^2}$

Squaring both sides, we get

${(x-6)}^2+{(y+6)}^2={(x-3)}^2+{(y-3)}^2$

$x^2-12x+36+y^2+12y+36=x^2-6x+9+y^2-6y+9$

$\Rightarrow -6x+18y+54=0$

or $x-3y=9$ ......$\left(1\right)$

Distance $AO=$Distance $BO$

Using distance formula,

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

Squaring both sides, we get

${(x-6)}^2+{(y+6)}^2={(x-3)}^2+{(y+7)}^2$

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

$3x+y=7$ .........$\left(2\right)$

Put $y=7-3x$ in $\left(1\right)$ we get

$x-3(7-3x)=9$

$\Rightarrow x-21+9x=9$

$\Rightarrow 10x=9+21=30$

$\Rightarrow x=dfrac3010=3$

$\therefore x=3$

Put $x=3$ in $y=7-3x=7-3\times 3=7-9=-2$

$\therefore x=3,y=-2$

$\therefore$ the centre of the circle is $(3,-2)$