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

**Solution:**

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula = √ [( x_{2 - }x_{1 })^{2} + (y_{2} - y_{1})^{2}]

Let's draw the required figure.

Given,

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

Let the points (6,- 6), (3, -7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.

Distance from centre O to A, B, C are found below using the Distance formula mentioned in the Reasoning.

Hence, OA = √ [(x - 6)^{2} + (y + 6)^{2}]

OB = √ [(x - 3)^{2} + (y + 7)^{2}]

OC = √ [(x - 3)^{2} + (y - 3)^{2}]

From the Figure that,

OA = OB (radii of the same circle)

√ [(x - 6)^{2} + (y + 6)^{2}] = √ [(x - 3)^{2} + (y + 7)^{2}]

x^{2} + 36 - 12x + y^{2} + 36 + 12y = x^{2} + 9 - 6x + y^{2} + 49 + 14y (Squaring on both sides)

- 6x - 2y + 14 = 0

3x + y = 7 .....Equation (1)

Similarly, OA = OC (radii of the same circle)

√ [(x - 6)^{2} + (y + 6)^{2}] = √ [(x - 3)^{2} + (y - 3)^{2}]

x^{2} + 36 - 12x + y^{2} + 36 + 12y = x^{2} + 9 - 6x + y^{2} + 9 - 6y (Squaring on both sides)

-6x + 18y + 54 = 0

-3x + 9y = - 27 .....Equation(2)

On adding Equation (1) and Equation (2), we obtain

10 y = - 20

y = - 2

From Equation (1), we obtain

3x - 2 = 7

3x = 9

x = 3

Therefore, the centre of the circle is (3, - 2).

**Video Solution:**

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

### Maths NCERT Solutions Class 10 - Chapter 7 Exercise 7.4 Question 3:

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

The centre of a circle passing through the points (6, - 6), (3, - 7) and (3, 3) is (3, - 2)