Find the centre of a circle passing through the points (6, − 6), (3, − 7) and (3, 3).
Solution
Let O (x, y) be the centre of the circle. And let the points (6, −6), (3, −7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.
`:.OA = sqrt((x-6)^2+(y+6)^2)`
`OB = sqrt((x-3)^2+(y+7)^2)`
`OC = sqrt((x-3)^2+(y-3)^2)`
However OA = OB (Radii of same circle)
`=>sqrt((x-6)^2+(y+6)^2)=sqrt((x-3)^2+(y+7)^2)`
=>x2+36 - 12x + y2 + 36 + 12y = x2 + 9 -6x + y2 + 49 -14y
⇒ -6x + 2y + 14 = 0
⇒ 3x + y = 7 ....1
Similary OA = OC (Radii of same circle)
`=sqrt((x-6)^2+(y+6)^2) = sqrt((x-3)^2 + (y -3)^2)`
=x2 + 36 - 12x +y2 + 36 + 12y = x2 + 9 - 6x + y2 + 9 - 6y
⇒ -6x + 18y + 54 = 0
⇒ -3x + 9y = -27 .....(2)
On adding equation (1) and (2), we obtain
10y = −20
y = −2
From equation (1), we obtain
3x − 2 = 7
3x = 9
x = 3
Therefore, the centre of the circle is (3, −2).