# 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 = √( x₂_{ - }x₁_{ })^{2} + (y₂ - y₁)^{2}

Let's draw the required figure.

According to the diagram,

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

Let the points (6,- 6), (3, -7), and (3, 3) represent 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.

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 ,

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 ..... (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 ..... (2)

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

3x + y - 3x + 9y = 7 + (-27)

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).

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 7

**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

**Summary:**

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

**☛ Related Questions:**

- The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of the other two vertices.
- The vertices of a ∆ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = 1/4.Calculate the area of the ∆ADE and compare it with the area of ∆ABC. (Recall Theorem 6.2 and Theorem 6.6).
- Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC.(i) The median from A meets BC at D. Find the coordinates of the point D.(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.(iv) What do yo observe? [Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]

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