Completing the Square in Circle Equations

Any equation of the form \(Ax^2 + Ay^2 + Bx + Cy + D = 0\) produces a circle — as long as both squared terms have the same coefficient. If the coefficients differ, you get an ellipse instead.

The problem is that this general form doesn't tell you much at a glance. To find the circle's center and radius, you need to rewrite it in standard form:

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

Here \((h, k)\) is the center and \(r\) is the radius. The technique for getting there is called completing the square.

How Completing the Square Works

Before applying it to circles, it's worth understanding what completing the square actually does.

When you expand a perfect square like \((x + 6)^2\), you get:

$$(x + 6)^2 = x^2 + 12x + 36$$

Notice the pattern: the constant (36) is always \(\left(\frac{\text{linear coefficient}}{2}\right)^2\). Here, \(\left(\frac{12}{2}\right)^2 = 6^2 = 36\). That's the whole trick — if you know the linear coefficient, you can figure out exactly what constant you'd need to make a perfect square.

So if you're given \(x^2 + 12x\) and want to turn it into a perfect square, take the linear coefficient (12), divide by 2, and square it: \(\left(\frac{12}{2}\right)^2 = 36\). Add 36 to both sides and the left side factors cleanly:

$$x^2 + 12x + 36 = 39$$ $$(x + 6)^2 = 39$$

That number you added (36) is the "magic number." You're not changing the equation — you're adding the same thing to both sides — but you've transformed the left side into something you can factor.

One important rule before you start: the coefficient on \(x^2\) must be 1. If it isn't, divide every term by that coefficient first. For example, if you have \(5x^2 - 2x - 4 = 0\), divide everything by 5 first to get \(x^2 - \frac{2}{5}x - \frac{4}{5} = 0\), then find your magic number from the new linear coefficient \(\left(\frac{1}{2} \cdot \frac{2}{5}\right)^2 = \frac{1}{25}\).

Applying It to a Circle Equation

Example: Rewrite \(4x^2 + 4y^2 - 24x + 32y - 4 = 0\) in standard form.

Step 1 — Check that the squared terms match. Both have a coefficient of 4, so this is a circle.

Step 2 — Divide every term by 4 so the squared terms have coefficient 1.

$$x^2 + y^2 - 6x + 8y - 1 = 0$$

Step 3 — Group the x terms and y terms, moving the constant to the right.

$$x^2 - 6x + y^2 + 8y = 1$$

Step 4 — Complete the square for x and y separately.

For the x group: \(\left(\frac{6}{2}\right)^2 = 9\). For the y group: \(\left(\frac{8}{2}\right)^2 = 16\).

Add both magic numbers to both sides — don't forget the right side:

$$(x^2 - 6x + 9) + (y^2 + 8y + 16) = 1 + 9 + 16$$

Step 5 — Factor each group.

$$(x - 3)^2 + (y + 4)^2 = 26$$

Reading Off the Center and Radius

Compare to standard form \((x - h)^2 + (y - k)^2 = r^2\):

  • The x part is \((x - 3)^2\), so \(h = 3\).
  • The y part is \((y + 4)^2\). Rewrite it as \((y - (-4))^2\), so \(k = -4\).
  • The right side is 26, so \(r = \sqrt{26}\).

Center: \((3, -4)\) — Radius: \(\sqrt{26}\)

The sign flip on the y coordinate is the most common place to make a mistake. When you see \((y + 4)^2\), the center's y-coordinate is \(-4\), not \(+4\). If the factored term has a plus sign, the coordinate is negative. If it has a minus sign, the coordinate is positive.

Try These

Work through each one and identify the center and radius.

1. \(x^2 + y^2 + 4x - 6y - 3 = 0\) Show answer\((x^2 + 4x + 4) + (y^2 - 6y + 9) = 3 + 4 + 9\), so \((x + 2)^2 + (y - 3)^2 = 16\). Center: \((-2, 3)\), Radius: \(4\).

2. \(x^2 + y^2 - 10x + 2y + 10 = 0\) Show answer\((x^2 - 10x + 25) + (y^2 + 2y + 1) = -10 + 25 + 1\), so \((x - 5)^2 + (y + 1)^2 = 16\). Center: \((5, -1)\), Radius: \(4\).

3. \(2x^2 + 2y^2 - 8x + 12y - 24 = 0\) Show answerDivide by 2 first: \(x^2 + y^2 - 4x + 6y - 12 = 0\). Then \((x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9\), so \((x - 2)^2 + (y + 3)^2 = 25\). Center: \((2, -3)\), Radius: \(5\).

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