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Derive the Equation
of a Circle

Learn how the Pythagorean Theorem connects a circle's center and radius to its algebraic equation, and master completing the square to reveal hidden circles.

Algebra 2

What Is a Circle, Really?

A circle is the set of all points in a plane that are the same distance from a fixed point. That fixed point is the center, and that constant distance is the radius.

\( C(h,k) \) \( r \) \( P(x,y) \) \( r \) \( r \)

From Distance to Equation

The distance formula measures the straight-line distance between two points. It comes directly from the Pythagorean Theorem:

If we set \( (x_1, y_1) = (h, k) \) (the center) and \( (x_2, y_2) = (x, y) \) (any point on the circle), and the distance equals \( r \):

The Pythagorean Connection

\( (h,k) \) \( (x,y) \) \( r \) \( (x-h) \) \( (y-k) \)
Pythagorean
Theorem:

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

The Standard Form

\( (h, k) \)
The center coordinates: the signs are opposite what appears inside the parentheses.
\( r^2 \)
The right side is the radius squared: take the square root to find \( r \).

Standard vs General Form

Standard Form General Form
Equation \( (x-h)^2 + (y-k)^2 = r^2 \) \( x^2 + y^2 + Dx + Ey + F = 0 \)
Center Read directly: \( (h, k) \) Hidden: must complete the square
Radius Read directly: \( \sqrt{r^2} \) Hidden: computed after rewriting
Best for Graphing and interpreting Expanded algebraic work

Watch the Signs Carefully

The standard form uses subtraction: \( (x - h) \). This means the sign you see inside the parentheses is opposite to the actual coordinate.

\( (x - 3)^2 \)
You see \( -3 \), so \( h = +3 \). The center's x-coordinate is positive 3.
\( (y + 5)^2 \)
This is really \( (y - (-5))^2 \), so \( k = -5 \). The center's y-coordinate is negative 5.

How to Write the Equation

Given a center and radius, follow these steps to write the equation in standard form:

1
Identify the center \( (h, k) \) and the radius \( r \).
2
Substitute into standard form: Replace \( h \), \( k \), and \( r \) in \( (x - h)^2 + (y - k)^2 = r^2 \).
3
Simplify the right side: Compute \( r^2 \) and leave the left side in factored form.
4
Double-check signs: Verify each parenthesis: \( (x - h) \) uses subtraction from \( h \).

Example 1: Write the Equation

Problem: Write the equation of a circle with center \( (2, -3) \) and radius \( r = 4 \).

\(x\) \(y\) \( (2,-3) \) \( r=4 \) O 2 −3

Example 2: Read Center and Radius

Problem: Find the center and radius of \( (x + 4)^2 + (y - 1)^2 = 49 \).

1
Rewrite \( (x + 4)^2 \) as \( (x - (-4))^2 \). So \( h = -4 \).
2
\( (y - 1)^2 \) gives \( k = 1 \) directly.
3
\( r^2 = 49 \), so \( r = 7 \).

Completing the Square

When a circle equation is in general form, you must complete the square to convert it to standard form. Here is the method:

1
Group \( x \)-terms and \( y \)-terms together. Move the constant to the right side.
2
For each group, take half the linear coefficient, square it, and add to both sides.
3
Factor each group as a perfect square binomial.
4
Read the center \( (h, k) \) and radius \( r \) from the result.

The Key Step: Half, Square, Add

The critical move in completing the square is:

Example: \( x^2 + 6x \)
Half of 6 is 3. Square of 3 is 9. Add 9:
\( x^2 + 6x + 9 = (x + 3)^2 \)
Example: \( y^2 - 10y \)
Half of \(-10\) is \(-5\). Square of \(-5\) is 25. Add 25:
\( y^2 - 10y + 25 = (y - 5)^2 \)

Example 3: Complete the Square

Problem: Find the center and radius of \( x^2 + y^2 - 6x + 4y - 12 = 0 \).

1
Group and move constant: \( (x^2 - 6x) + (y^2 + 4y) = 12 \)
2
Complete the square for \( x \): Half of \(-6\) is \(-3\), squared is 9. Add 9 to both sides.
3
Complete the square for \( y \): Half of \( 4 \) is \( 2 \), squared is 4. Add 4 to both sides.
4
Result: \( (x - 3)^2 + (y + 2)^2 = 12 + 9 + 4 = 25 \)

Example 4: Trickier Completion

Problem: Rewrite \( x^2 + y^2 + 8x - 2y + 8 = 0 \) in standard form.

1
Group: \( (x^2 + 8x) + (y^2 - 2y) = -8 \)
2
x-group: \( \left(\frac{8}{2}\right)^2 = 16 \). Add 16 to both sides.
3
y-group: \( \left(\frac{-2}{2}\right)^2 = 1 \). Add 1 to both sides.
4
Factor: \( (x + 4)^2 + (y - 1)^2 = -8 + 16 + 1 = 9 \)

Common Mistakes

✕ Forgetting to Flip Signs
Reading \( (x + 4)^2 \) as center \( h = 4 \) instead of \( h = -4 \). Always flip the sign inside the parenthesis.
✕ Not Adding to Both Sides
Adding the completing-the-square value to the left side only. Whatever you add to the left, you must also add to the right.
✕ Using \( r \) Instead of \( r^2 \)
Writing \( r^2 = 5 \) and concluding \( r = 5 \). Remember: the equation gives \( r^2 \), so take the square root.

Summary at a Glance

Task Key Action Result
Write equation from center & radius Plug into \( (x-h)^2+(y-k)^2=r^2 \) Standard form
Read center from standard form Flip signs inside parentheses \( (h, k) \)
Read radius from standard form Take \( \sqrt{\text{right side}} \) \( r \)
Convert general → standard Complete the square (both \( x \) and \( y \)) Center and radius revealed

Question 1 of 5

What is the equation of a circle with center \( (5, -1) \) and radius \( r = 3 \)?

Question 2 of 5

What are the center and radius of the circle \( (x + 7)^2 + (y - 4)^2 = 36 \)?

Question 3 of 5

To complete the square on \( x^2 - 10x \), what value must be added?

Question 4 of 5

Find the center and radius of the circle given by
\( x^2 + y^2 + 12x - 4y + 15 = 0 \).

Question 5 of 5

A circle is centered at the origin with radius \( r = \sqrt{10} \). Which equation represents this circle?