4x2 + 4y2 – 24x + 16y – 100 = 0 into standard circle form

Question

Show steps please, need help. or even just show me an example of an equation that is similar.

but please keep both the 4x^2 and 4y^2 if you show an example.

4x^2 + 4y^2 – 24x + 16y – 100 = 0

Answer 1

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.