Question

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

Answer 1

A = (6, -6), B = (3, -7), C = (3, 3)
If O is the centre, then OA = OB = OC (radii are equal)
If O = (x, y), then
$OA=\sqrt{(x-6{)}^2+(y+6{)}^2}$
$OB=\sqrt{(x-3{)}^2+(y+7{)}^2}$
$OC=\sqrt{(x-3{)}^2+(y-3{)}^2}$
Now, OA =OB
$\Rightarrow (x-6{)}^2+(y+6{)}^2=(x-3{)}^2+(y+7{)}^2$
$\Rightarrow x^2-12x+36+y^2+12y+36=x^2-6x+9+y^2+14y+49$
$\Rightarrow x^2-12x+y^2+12y+72=x^2-6x+y^2+14y+58$
$\Rightarrow x^2-12x-(x^2-6x)=y^2+14y-(y^2+12y)+58-72$
$\Rightarrow -12x+6x=14y-12y-14$
$\Rightarrow -6x=2y-\mathrm{14.....}\left(1\right)$

Similarly OB=OC,
$(x-3{)}^2+(y+7{)}^2$
$=(x-3{)}^2+(y-3{)}^2$
$\Rightarrow (x-3{)}^2-(x-3{)}^2+(y+7{)}^2$
$=(y-3{)}^2$
$\Rightarrow y^2+14y+49=y^2-6y+9$
$\Rightarrow 20y=-40$
$\Rightarrow y=-2$
Substituting the value of y in equation (1), we get;
$-6x=2y-14$
$\Rightarrow -6x=-4-14=-18$
$\Rightarrow x=3$
Hence, x = 3, y = - 2 are the coordinates of centre.