Finding the center and radius from a circle equation

Expert intro: A circle equation in general form becomes easy to read once the x-terms and y-terms are completed into perfect squares.

Detailed walkthrough

Identify the circle form

Start with

$x^2+y^2+10x+8y-40=0.$

Move the constant to the other side:

$x^2+10x+y^2+8y=40.$

Now complete the square for x and y separately.

Complete the square

For the x-terms,

$x^2+10x=(x+5)^2-25.$

For the y-terms,

$y^2+8y=(y+4)^2-16.$

Substitute these into the equation:

$(x+5)^2-25+(y+4)^2-16=40.$

Combine constants:

$(x+5)^2+(y+4)^2=81.$

Read the center and radius

A circle in standard form is

$(x-h)^2+(y-k)^2=r^2.$

So here,

• center: $(-5,-4)$
• radius: $\sqrt{81}=9$

Why the given work is correct

The equation has been rewritten into standard circle form, and the center comes from the signs inside each squared binomial. The radius is the square root of the constant on the right-hand side.

Quick check

Expanding $(x+5)^2+(y+4)^2=81$ gives back

$x^2+y^2+10x+8y-40=0,$

so the result is consistent.

💡 Pitfall guide

A common error is to read the center from the middle signs instead of the signs inside the parentheses. Since the standard form is $(x-h)^2+(y-k)^2=r^2,$ the center is $(-5,-4)$, not $(5,4)$. Another mistake is forgetting that the radius is the square root of the right-hand side, not the right-hand side itself. Here the radius is 9, not 81.

🔄 Real-world variant

If the equation were

$x^2+y^2-6x+14y+49=0,$

you would move the constant, complete the square, and get

$(x-3)^2+(y+7)^2=9,$

so the center would be $(3,-7)$ and the radius would be 3. The method does not change; only the signs and constants do. That is why completing the square is the standard tool for every circle in general form.

🔍 Related terms

completing the square, standard form of a circle, circle center and radius