Question

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

Answer 1

The distance d between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula

`d = sqrt((x_1 - x_2)^2 + (y_1 -  y_2)^2)` 

The centre of a circle is at equal distance from all the points on its circumference.

Here it is given that the circle passes through the points A(6,−6), B(3,−7) and C(3,3).

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

So we have AO = BO = CO

`AO = sqrt((6 - x)^2 + (-6-y)^2)`

`BO  = sqrt((3 - x)^2 + (-7 - y)^2)`

`CO = sqrt((3 - x)^2 + (3 - y)^2)`

Equating the first pair of these equations we have,

AO = BO

`sqrt((6 - x)^2 + (-6-y)^2) = sqrt((3 -x)^2 + (3 -y)^2)`

Squaring on both sides of the equation we have,

`(6 - x)^2 + (-6-y)^2 = (3 - x)^2 + (3 - y)^2`

`36 + x^2 - 12x + y^2 + 12y = 9 + x^2 - 6x + y^2 - 6y`

6x - 18y = 54

`x - 3y= 9`

Now we have two equations for ‘x’ and ‘y’, which are

3x + y = 7

x - 3y = 9

From the second equation we have y = 3x + 7. Substituting this value of ‘y’ in the first quation we have,

`x - 3(-3x + 7) = 9`

x + 9x - 21 = 9

10x = 30

x = 3

Therefore the value of ‘y’ is,

y = 3x + 7

= -3(3) + 7

y = -2

Hence the co-ordinates of the centre of the circle are (3, -2).