Find the equation of the circle passes through the points `(2, 1)` and `(-2, 3)` and has centre on the line `x + y + 4 = 0`

Explanation

Let the equation of the circle be `(x-h)^(2) + (y-k)^(2) = r^(2)`
Since the circle passes through `(2, 1)` and `(-2, 3)`
We have, `(2-h)^(2) + (1-k)^(2) = r^(2)" " …(1)`
and `(-2-h)^(2) + (3-k)^(2) = r^(2)" "...(2)`
SInce the centre lies on the line `x + y + 4 = 0`
`therefore " " We have, `h + k + 4 =0` "...(3)`
Equation (1) and (2) we get :
`(2-h)^(2) + (1-k)^(2) = (-2-h)^(2) + (3-k)^(2)`
`rArr 2h - k + 2=0`
Solving (4) and (3), we get
`h = -2` and `k =-2`
Now, putting the values of h and k in (1), we get,
`[2-(2)]^(2) + [1-(-2)]^(2) = r^(2)`
`rArr 16 + 9 = r^(2)`
`rArr r = pm 5`
The value of r cannot be negative,
`therefore r = 5, (h,k) = (-2, -2)`
Hence, equation of the required circle is `[x-(-2)]^(2) + [y-(-2)]^(2) = (5)^(2)`
i.e., `(x+2)^(2) + (y+2)^(2) = (5)^(2)`