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Solution :

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