# Find the Centre of a Circle Passing Through the Points (6, − 6), (3, − 7) and (3, 3). - Mathematics

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

#### Solution

Let O (xy) 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).

Concept: Area of a Triangle
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#### APPEARS IN

NCERT Class 10 Maths
Chapter 7 Coordinate Geometry
Exercise 7.4 | Q 3 | Page 171