Question

Find the center of a circle passing through the points ( 6,-6), ( 3, -7) and (3,3).

Answer 1

Let $(x_1,y_1)$ be the centre of the given circle that passes through the points (6-6), (3, -7) and (3,3)
The distance of all these points from the centre should be equals to the radius of the circle
$\therefore (x_1-6{)}^2+(y_1,+6{)}^2=r^2$ ....(1)
$(x_1-3{)}^2+(y_1+7{)}^2=r^2$ ....(2)
$(x_1-3{)}^2+(y_1-3{)}^2=r^2$ .....(3)
from (2) $\&$ (3)
$(\text{/}{x}_1-3{)}^2+(y_1+7{)}^2=(\text{/}{x}_1-3{)}^2+(y_1-3{)}^2\Rightarrow \text{/}{y_1}^2+14y_1+49=\text{/}{y_1}^2-6y_1+9\Rightarrow 20y_1=-40\Rightarrow y_1=-2$
From (1) $\&$ 2
$(x_1-6{)}^2+(y_1+6{)}^2=(x_1-3{)}^2+(y_1+7{)}^2$
putting $y_1=-2$
$(x_1-6{)}^2+(-2+6{)}^2=(x_1-3{)}^2+(-2+7{)}^2$
$\Rightarrow (x_1-6{)}^2+(y_1+6{)}^2=(x_1-3{)}^2+(y_1+7{)}^2$
putting $y_1=-2$
$\Rightarrow (x_1-6{)}^2+(-2+6{)}^2=(x_1-3{)}^2+(-2+7{)}^2\Rightarrow \left(\text{/}{x_1}^2\right)-12x_1+36+16=\text{/}{x}_1^2-6x_1+9+25\Rightarrow -6x_1=-18\Rightarrow x_1=3$
$\therefore$ The centre of the circle is (3,-2 )