Question

Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).

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(5,−8), B(2,−9) and C(2,1).

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

So we have AO =  BO = CO

`AO = sqrt((5 - x)^2 + (-8 - y)^2)`

`BO = sqrt((2 - x)^2 + (-9-y)^2)`

`CO = sqrt((2 - x)^2 + (1 - y)^2)`

Equating the first pair of these equations we have,

AO = BO

`sqrt((5 - x)^2 + (-8 - y)^2) = sqrt((2 - x)^2 + (-9 - y)^2)`

Squaring on both sides of the equation we have,

`(5 -x)^2 + (-8 - y)^2 = (2 - x)^2 + (-9 - y)^2`

`25 + x^2 - 10x + 64 + y^2 + 16y = 4 + x^2 - 4x + 81 + y^2 + 18y` 

6x + 2y = 4

3x + y = 2

Equating another pair of the equations we have,

AO =BO

`sqrt((5 - x)^2 + (-8 - y)^2) = sqrt((2 - x)^2 + (-9-y)^2)`

Squaring on both sides of the equation we have,

`(5 - x)^2 + (-8 - y)^2 = (2 - x)^2 + (-9 - y)^2`

`25 + x^2 - 10x + 64 + y^2 + 16y = 4 + x^2 - 4x + 81 + y^2 + 18y`

6x + 2y = 4

3x + y = 2

Equating another pair of the equations we have,

AO = CO

`sqrt((5 -x)^2 + (-8 - y)^2) = sqrt((2 -x)^2 + (1 - y)^2)`

`25 + x^2 - 10x + 64 + y^2 + 16y = 4 + x^2 -4x + 1 + y^2 - 2y`

6x - 8y = 84

x - 3y = 14

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

3x + y = 2

x - 3y = 14

From the second equation we have y = -3x + 2. Substituting this value of ‘y’ in the first equation we have,

x - 3(-3x + 2) = 14

x + 9x - 6 = 14

10x = 20

x = 2

Therefore the value of ‘y’ is,

y = -3x + 2

= -3(2) + 2

y = -4

Hence the co-ordinates of the centre of the circle are (2, -4)