We need to complete the square for x and y.
4x2-24x can be written: 4(x2-6x) and 4y2+16y can be written: 4(y2+4y).
In each of the parentheses we need to add a constant to make a couple of perfect squares. To find the constants we halve the x and y coefficients and then square them. For x we halve 6 to get 3 then square it to get 9. Similarly for y we get 2 then 22=4.
So we arrive at: 4(x2-6x+9) and 4(y2+4y+4). But we need to make allowance for the two constants. For x we added 4×9=36 and for y we added 4×4=16. So we added 36+16=52 to the expression. Therefore we must subtract 52 to balance out what was added. So the original equation becomes:
4(x2-6x+9)+4(y2+4y+4)-52-100=0.
This then becomes 4(x-3)2+4(y+2)2=152, replacing the expanded perfect squares by their square roots squared. Finally, to get the standard form we need to divide through by 4:
(x-3)2+(y+2)2=38.
In this form we can see immediately that the centre of the circle is at (3,-2). The radius is √38.