Question

The centre of the circle passing through the points (8,12),(11,3) and (0,14) is

Answer 1

Given:

The points are A(8, 12) , B( 11, 3) , C(0 ,14 ).

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

So, the distance of each point from centre will be equal.

OA = OB = OC

`\begin{gathered} OA=OB \\ \sqrt[]{(x-8)^2+(y-12)^2}=\sqrt[]{(x-11)^2+(y-3)^2} \\ (x-8)^2+(y-12)^2=(x-11)^2+(y-3)^2 \\ y^2-24y+x^2+208-16x=y^2-6y+x^2+130-22x \\ x^2+208-16x=18y+x^2-22x+130 \\ x^2-16x=x^2-22x+18y-78 \\ 6x=18y-78 \\ x-3y=-13\ldots\ldots\ldots\ldots\text{.}(1) \end{gathered}`

Now,

`\begin{gathered} OB=OC \\ \sqrt[]{(x-11)^2+(y-3)^2}=\sqrt[]{(x-0)^2+(y-14)^2} \\ (x-11)^2+(y-3)^2=(x-0)^2+(y-14)^2 \\ y^2-6y+x^2+130-22x=x^2+y^2-28y+196 \\ x^2+130-22x=-22y+x^2+196 \\ x^2+130-22x=-22y+x^2+196 \\ -22x=66-22y \\ y-x=3\ldots\ldots\ldots\text{.}.(2) \end{gathered}`

Solve the equation (1) and (2).

`\begin{gathered} x-3y=-13\ldots\ldots\ldots(1) \\ y-x=3\ldots\ldots\text{.}(2) \\ \text{equation (2) can be written as,} \\ x=y-3\text{ put it in equation 1,} \\ y-3-3y=-13 \\ -2y=-10 \\ y=5 \\ \text{equation (2) becomes,} \\ 5-x=3 \\ 5-3=x \\ x=2 \\ \text{Center= O(x,y)=(2,5)} \end{gathered}`

Answer: center of circle is ( 2, 5 ).