Find the radius and centre of the circle passing through the points `(2,0), (6, 0)` and `(4, 2)`.

Explanation

The general equation of the circle is given by `x^(2) + y^(2) + 2gx + 2fy + c =0" "…(1)`
where radius `= sqrt(g^(2) + f^(2) -c)` and centre `=(-g, -f)`
`therefore` To find the required radius and centre, we first need to find the equation of the circle passing through the three given points.
Since, the circle passes through the point (2, 0), (6, 0) and (4, 2) these points will satisfy equation (1)
i.e., `(2)^(2) + 0 + 2g(2) + 2f(0) + c = 0`,
`(6)^(2) + 0 + 2g(6) + 2f(0) + c =0`
and `(4)^(2) + (2)^(2) + 2g(4) + 2f(2) + c = 0`
`rArr 4g + c + 4 = 0" "...(2)`,
`12g + c + 36 = 0 and " "...(3)`
`4f + 8g + c + 20 = 0" "...(4)`
Equation (2) and (3) given g = -4 and c = 12
Putting these value in (4) we get : f = 0
Now, radius and centre of the given circle is given by `sqrt(g^(2) + f^(2) -c)` and `(-g, -f)` respectively.
i.e., radius = `sqrt((-4)^(2) + (0)^(2) - 12 ) = 2` and centre = `(-g, -f)` i.e., (4, 0)