Circles

A circle is the set of all points in a plane that are equidistant from a fixed point called the center. That constant distance is the radius. Circles are the simplest conic section and the starting point for understanding ellipses, hyperbolas, and parabolas. In this lesson you will learn to write the equation of a circle in standard form, convert from general form by completing the square, graph circles, and find equations from geometric information.

Standard Form of a Circle

The standard form of a circle with center $(h, k)$ and radius $r$ is:

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

This equation says: any point $(x, y)$ on the circle is exactly $r$ units from the center $(h, k)$. It follows directly from the distance formula.

Special case: When the center is at the origin $(0, 0)$:

$x^2 + y^2 = r^2$

Reading Center and Radius from Standard Form

Given $(x - 3)^2 + (y + 2)^2 = 25$:

• Center: $(3, -2)$ — notice the signs flip: $(x - 3)$ means $h = 3$, and $(y + 2) = (y - (-2))$ means $k = -2$
• Radius: $r = \sqrt{25} = 5$

General Form of a Circle

Expanding the standard form gives the general form:

$x^2 + y^2 + Dx + Ey + F = 0$

This form is less immediately useful because the center and radius are hidden. You need to complete the square to convert back to standard form.

Converting General Form to Standard Form

This is the most important skill in this lesson. The process uses completing the square on both the $x$ and $y$ terms.

Worked Example 1: Convert to Standard Form

Convert $x^2 + y^2 - 6x + 4y - 12 = 0$ to standard form and find the center and radius.

Step 1 — Group $x$ terms and $y$ terms, and move the constant:

$x^2 - 6x + y^2 + 4y = 12$

Step 2 — Complete the square for $x$:

Half of $-6$ is $-3$. Square it: $(-3)^2 = 9$. Add 9 to both sides.

$x^2 - 6x + 9 + y^2 + 4y = 12 + 9$

Step 3 — Complete the square for $y$:

Half of $4$ is $2$. Square it: $2^2 = 4$. Add 4 to both sides.

$x^2 - 6x + 9 + y^2 + 4y + 4 = 12 + 9 + 4$

Step 4 — Factor the perfect square trinomials:

$(x - 3)^2 + (y + 2)^2 = 25$

Result: Center $(3, -2)$, radius $r = \sqrt{25} = 5$.

Worked Example 2: A Circle with Fractional Coefficients

Convert $x^2 + y^2 + 3x - 5y + 2 = 0$ to standard form.

Step 1: $x^2 + 3x + y^2 - 5y = -2$

Step 2 (complete $x$): Half of $3$ is $\frac{3}{2}$. Square: $\frac{9}{4}$.

$x^2 + 3x + \frac{9}{4} + y^2 - 5y = -2 + \frac{9}{4}$

Step 3 (complete $y$): Half of $-5$ is $-\frac{5}{2}$. Square: $\frac{25}{4}$.

$x^2 + 3x + \frac{9}{4} + y^2 - 5y + \frac{25}{4} = -2 + \frac{9}{4} + \frac{25}{4}$

Step 4 — Simplify the right side and factor:

$\left(x + \frac{3}{2}\right)^2 + \left(y - \frac{5}{2}\right)^2 = -2 + \frac{34}{4} = -\frac{8}{4} + \frac{34}{4} = \frac{26}{4} = \frac{13}{2}$

Result: Center $\left(-\frac{3}{2}, \frac{5}{2}\right)$, radius $r = \sqrt{\frac{13}{2}} = \frac{\sqrt{26}}{2} \approx 2.55$.

Writing the Equation from Geometric Information

From Center and Radius

If the center is $(-1, 4)$ and the radius is 3:

$(x + 1)^2 + (y - 4)^2 = 9$

From Endpoints of a Diameter

If the endpoints of a diameter are $A = (2, -1)$ and $B = (8, 5)$, then:

Center = midpoint of $AB$:

$h = \frac{2 + 8}{2} = 5, \quad k = \frac{-1 + 5}{2} = 2$

Center: $(5, 2)$.

Radius = half the diameter length:

$d = \sqrt{(8-2)^2 + (5-(-1))^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$

$r = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$

Equation:

$(x - 5)^2 + (y - 2)^2 = (3\sqrt{2})^2 = 18$

From Center and a Point on the Circle

If the center is $(1, -3)$ and the circle passes through $(4, 1)$:

$r = \sqrt{(4-1)^2 + (1-(-3))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Equation: $(x - 1)^2 + (y + 3)^2 = 25$

Graphing a Circle

To graph a circle from its equation:

1. Get standard form (complete the square if needed)
2. Plot the center $(h, k)$
3. From the center, count $r$ units up, down, left, and right to find four points on the circle
4. Sketch the curve through those four points

For the circle $(x - 3)^2 + (y + 2)^2 = 25$: plot center $(3, -2)$, then mark $(8, -2)$, $(-2, -2)$, $(3, 3)$, and $(3, -7)$. Connect with a smooth curve.

Real-World Application: Carpentry

A carpenter is designing a circular tabletop. The table must fit inside a rectangular alcove that is 6 feet wide. The center of the table will be placed at coordinates $(3, 3)$ relative to the corner of the alcove.

The maximum radius is 3 feet (half of 6 feet). To verify the table fits, the carpenter writes the equation:

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

Checking the leftmost point: $x = 3 - 3 = 0$ (touches the left wall). Rightmost: $x = 3 + 3 = 6$ (touches the right wall). The table fits exactly.

If the carpenter wants 1 foot of clearance on each side, the usable radius is $3 - 1 = 2$ feet, and the equation becomes:

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

When the Equation Does Not Represent a Circle

After completing the square, if the right side is zero, you get a single point (a “degenerate circle”). If the right side is negative, there is no graph — no real solutions exist.

Example: $x^2 + y^2 - 2x + 4y + 5 = 0$

$(x - 1)^2 + (y + 2)^2 = -5 + 1 + 4 = 0$

This is the single point $(1, -2)$.

Example: $x^2 + y^2 - 2x + 4y + 8 = 0$

$(x - 1)^2 + (y + 2)^2 = -8 + 1 + 4 = -3$

No real graph exists — this equation has no solution.

Common Mistakes

1. Forgetting to add the completed-square constant to both sides. If you add $9$ to the left, you must add $9$ to the right.
2. Sign errors when reading center from standard form. $(x + 3)^2$ means $h = -3$, not $+3$.
3. Confusing $r^2$ with $r$. If the equation says $= 16$, the radius is $4$, not $16$.
4. Leaving a leading coefficient on $x^2$ or $y^2$. If the equation is $2x^2 + 2y^2 + \ldots = 0$, divide everything by 2 first.

Practice Problems

Key Takeaways

• The standard form of a circle is $(x - h)^2 + (y - k)^2 = r^2$ with center $(h, k)$ and radius $r$
• Complete the square on both $x$ and $y$ terms to convert from general form to standard form
• The center coordinates have opposite signs from what appears in the equation: $(x + 3)^2$ gives $h = -3$
• To find the equation from diameter endpoints, use the midpoint for center and half the distance for radius
• If the right side of standard form is zero, the “circle” is a single point; if negative, no graph exists
• Circles are a special case of ellipses where $a = b$ (equal semi-axes)

Return to College Algebra for more topics in this section.