Deriving Circle Equations- Complete Tutorial

What Is a Circle Equation?

A circle equation describes every point that sits at a fixed distance from a center point. That fixed distance is the radius. The equation gives you a way to check if any point (x, y) belongs to the circle.

There are two forms you need to know:

• Standard form: (x - h)² + (y - k)² = r²
• General form: x² + y² + Dx + Ey + F = 0

The standard form is what you use when you know the center and radius. The general form is what you get after expanding and rearranging.

Deriving the Circle Equation from the Definition

Here's the logic. A circle is set of all points equidistant from a center point. Pick a center at (h, k) and a radius of r. Take any point (x, y) on the circle.

The distance from (x, y) to (h, k) must equal r. Use the distance formula:

√[(x - h)² + (y - k)²] = r

Square both sides to eliminate the square root. You get:

(x - h)² + (y - k)² = r²

That's it. That's the circle equation in standard form.

Special Case: Circle Centered at the Origin

If the center sits at (0, 0), then h = 0 and k = 0. Plug those in:

x² + y² = r²

This is the simplest version. The center is the origin, so you don't subtract anything from x or y.

How to Write the Equation When You Know Center and Radius

Step 1: Identify the center coordinates (h, k)

Step 2: Identify the radius r

Step 3: Plug into (x - h)² + (y - k)² = r²

Example: Center at (3, -2), radius = 5

(x - 3)² + (y - (-2))² = 5²

(x - 3)² + (y + 2)² = 25

Notice the sign flip. The equation uses (x - h), so if h = 3, you write (x - 3). If k = -2, you write (y - (-2)) which becomes (y + 2).

Converting Between Standard and General Form

Sometimes you need the general form for graphing calculators or algebraic manipulation.

To convert from standard to general:

Expand the squares and collect terms.

Example: (x - 3)² + (y + 2)² = 25

Expand:

x² - 6x + 9 + y² + 4y + 4 = 25

Collect:

x² + y² - 6x + 4y + 13 - 25 = 0

x² + y² - 6x + 4y - 12 = 0

That's the general form. You can check by completing the square in reverse.

Converting General Form to Standard Form

When you have a circle in general form, complete the square for x and y separately.

Example: x² + y² - 4x + 6y - 3 = 0

Group x terms and y terms:

(x² - 4x) + (y² + 6y) = 3

Complete the square for each group:

For x² - 4x: take half of -4, square it = 4

For y² + 6y: take half of 6, square it = 9

Add these to both sides:

(x² - 4x + 4) + (y² + 6y + 9) = 3 + 4 + 9

Factor:

(x - 2)² + (y + 3)² = 16

Center is (2, -3), radius is 4.

How to Find the Center and Radius from an Equation

Look at the standard form: (x - h)² + (y - k)² = r²

The center comes from the signs inside the parentheses. The right side tells you r², so take the square root for r.

Example: (x + 5)² + (y - 1)² = 49

Rewrite with minus signs: (x - (-5))² + (y - 1)² = 49

Center: (-5, 1)

Radius: √49 = 7

That's all there is to it. If you see a plus sign in the parentheses for x, the h value is negative. If you see a minus sign, h is positive.

Common Mistakes to Avoid

• Forgetting to flip the sign when reading the center. (x - 3) means the x-coordinate of the center is 3, not -3.
• Confusing r² with r. The right side of the standard form is r². Take the square root to get the actual radius.
• Not completing the square correctly. Remember to add the same value to both sides of the equation.
• Leaving the equation unsimplified when the problem asks for a specific form.

Circle Equation Forms Comparison

Worked Example: Writing an Equation from Three Points

Sometimes you don't get the center directly. You get three points on the circle and need to find the equation.

Problem: Find the equation of the circle passing through (2, 1), (4, 5), and (1, 2).

Set up the general form x² + y² + Dx + Ey + F = 0 and substitute each point.

For (2, 1): 4 + 1 + 2D + E + F = 0 → 2D + E + F = -5

For (4, 5): 16 + 25 + 4D + 5E + F = 0 → 4D + 5E + F = -41

For (1, 2): 1 + 4 + D + 2E + F = 0 → D + 2E + F = -5

Solve the system. Subtract the first equation from the third:

(D + 2E + F) - (2D + E + F) = -5 - (-5)

-D + E = 0 → E = D

Subtract the first from the second:

(4D + 5E + F) - (2D + E + F) = -41 - (-5)

2D + 4E = -36 → D + 2E = -18

Since E = D, substitute: D + 2D = -18 → 3D = -18 → D = -6

Then E = -6

Plug back: 2(-6) + (-6) + F = -5 → -12 - 6 + F = -5 → F = 13

Equation: x² + y² - 6x - 6y + 13 = 0

Convert to standard form by completing the square:

(x² - 6x) + (y² - 6y) = -13

(x - 3)² - 9 + (y - 3)² - 9 = -13

(x - 3)² + (y - 3)² = 5

Center: (3, 3), radius: √5

Quick Reference: Derivation Steps

Deriving from distance formula:

1. Start with distance = √[(x₂ - x₁)² + (y₂ - y₁)²]
2. Set distance = r
3. Substitute center (h, k) for (x₁, y₁)
4. Substitute point (x, y) for (x₂, y₂)
5. Square both sides
6. Result: (x - h)² + (y - k)² = r²

Deriving from Pythagorean theorem works the same way. The radius forms a right triangle with the x and y differences from the center to any point on the circle.

When the Equation Is Not a Circle

Sometimes completing the square gives you nonsense:

(x - 2)² + (y + 1)² = -9

The radius squared is negative. That's impossible for a real circle. This is an imaginary circle — no real points satisfy the equation.

(x - 2)² + (y + 1)² = 0

This is a point circle. The radius is 0. The "circle" is just a single point at (2, -1).

Check your right-side value. If it's positive, you have a real circle. If it's zero, you have a point. If it's negative, you have no real locus.