The Standard Form

(x - h)² + (y - k)² = r²

(h, k) = Center of the circle

r = Radius of the circle

📝 Derivation from Distance Formula

The distance from any point (x, y) on the circle to the center (h, k) is r:

r = √[(x - h)² + (y - k)²]

Squaring both sides gives the standard form equation.

The General Form

x² + y² + Dx + Ey + F = 0

D, E, F = Constants

🔹 Relationship to Standard Form

From general form, we can find:

h = -D/2

k = -E/2

r = √(h² + k² - F)

From Standard Form

Given: (x - h)² + (y - k)² = r²

• Center = (h, k)

• Radius = r = √(r²)

⚠️ Note: Watch the signs! (x - 3) means h = 3, but (x + 3) means h = -3

📝 Example 1

Find the center and radius of: (x - 4)² + (y + 2)² = 25

Solution:

• (x - 4) → h = 4

• (y + 2) = (y - (-2)) → k = -2

• r² = 25 → r = 5

Center: (4, -2), Radius: 5

📝 Example 2

Find the center and radius of: x² + y² = 16

Solution:

This is (x - 0)² + (y - 0)² = 16

Center: (0, 0) [Origin], Radius: 4

Method: Completing the Square

Steps:

Step 1: Group x terms and y terms together

Step 2: Move the constant to the right side

Step 3: Complete the square for x and y separately

Step 4: Add the same values to both sides

Step 5: Factor to get standard form

💡 Completing the Square Formula

For x² + bx, add (b/2)² to complete the square:

x² + bx + (b/2)² = (x + b/2)²

📝 Example

Convert to standard form: x² + y² + 6x - 4y - 3 = 0

Solution:

Step 1: Group terms: (x² + 6x) + (y² - 4y) = 3

Step 2: Complete the square for x: (6/2)² = 9

Step 3: Complete the square for y: (-4/2)² = 4

Step 4: Add to both sides: (x² + 6x + 9) + (y² - 4y + 4) = 3 + 9 + 4

Step 5: Factor: (x + 3)² + (y - 2)² = 16

Answer: Center (-3, 2), Radius 4

Method 1: Given Center and Radius

Simply plug the values into the standard form:

(x - h)² + (y - k)² = r²

📝 Example 1

Write the equation of a circle with center (2, -5) and radius 6.

Solution:

h = 2, k = -5, r = 6

(x - 2)² + (y + 5)² = 36

Method 2: Given Center and a Point on Circle

Step 1: Use the distance formula to find radius:

r = √[(x₂ - x₁)² + (y₂ - y₁)²]

Step 2: Plug center and radius into standard form

📝 Example 2

Write the equation of a circle with center (1, 3) that passes through (4, 7).

Solution:

Step 1: Find radius: r = √[(4-1)² + (7-3)²] = √[9 + 16] = √25 = 5

Step 2: Write equation with h = 1, k = 3, r = 5

(x - 1)² + (y - 3)² = 25

Method: Substitute and Check

To test if point (a, b) lies on the circle:

Step 1: Substitute x = a and y = b into the circle equation

Step 2: Simplify both sides

Step 3: If the equation is TRUE, the point is ON the circle

📝 Alternative: Distance Check

Calculate distance from point to center:

• If distance = r → Point is ON the circle

• If distance < r → Point is INSIDE the circle

• If distance > r → Point is OUTSIDE the circle

📝 Example

Does point (5, 2) lie on the circle (x - 2)² + (y - 6)² = 25?

Solution:

Substitute x = 5, y = 2:

(5 - 2)² + (2 - 6)² = 25

3² + (-4)² = 25

9 + 16 = 25

25 = 25 ✓ TRUE

Yes, the point (5, 2) lies ON the circle.

From Standard Form

Steps:

Step 1: Identify the center (h, k) from (x - h)² + (y - k)²

Step 2: Find the radius r from r²

Step 3: Plot the center point on the coordinate plane

Step 4: From the center, count r units in all four directions (up, down, left, right)

Step 5: Connect these points in a smooth circle

From General Form

Steps:

Step 1: Convert to standard form by completing the square

Step 2: Identify center and radius

Step 3: Graph using the standard form method

📝 Example

Graph: (x + 1)² + (y - 2)² = 9

Solution:

• Center: (-1, 2)

• Radius: √9 = 3

• Plot center at (-1, 2)

• Mark points: (-4, 2), (2, 2), (-1, 5), (-1, -1)

• Draw circle through these points

The Connection

A circle is defined as all points at distance r from center (h, k).

For any point (x, y) on the circle, we form a right triangle:

• Horizontal leg: |x - h|

• Vertical leg: |y - k|

• Hypotenuse: r (radius)

Derivation

By Pythagorean Theorem:

(x - h)² + (y - k)² = r²

This is the standard form of the circle equation!

🔹 Circle Equation Forms

🔹 Key Formulas