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Equations of a Circle

CSEC Additional Mathematics Essential Knowledge: A circle is the set of all points in a plane that are equidistant from a fixed point called the center. The distance from the center to any point on the circle is the radius. Understanding circle equations is crucial for coordinate geometry and has applications in physics, engineering, and computer graphics.

Key Concept: The standard equation of a circle with center at point \((h, k)\) and radius \(r\) is \((x – h)^2 + (y – k)^2 = r^2\). The general form of a circle equation is \(x^2 + y^2 + 2gx + 2fy + c = 0\), where the center is \((-g, -f)\) and radius is \(\sqrt{g^2 + f^2 – c}\).

Part 1: Standard Form of Circle Equation

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The Fundamental Equation

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Distance Formula Connection

A circle is defined as all points \((x, y)\) that are a fixed distance \(r\) (radius) from a fixed point \((h, k)\) (center). Using the distance formula:

\[\sqrt{(x – h)^2 + (y – k)^2} = r\]

Squaring both sides gives us the standard form:

\[(x – h)^2 + (y – k)^2 = r^2\]

Center: (h, k)
Radius: r
Equation:
\((x-h)^2 + (y-k)^2 = r^2\)

Special Case: Center at Origin

When the center is at \((0, 0)\), the equation simplifies to:

\(x^2 + y^2 = r^2\)

Identifying Components

In \((x – h)^2 + (y – k)^2 = r^2\):

\(h\) = x-coordinate of center
\(k\) = y-coordinate of center
\(r\) = radius (always positive)

Signs Matter!

The equation has MINUS signs: \((x – h)^2 + (y – k)^2 = r^2\)

If the center is \((-3, 4)\), then \(h = -3\), \(k = 4\), and the equation becomes:
\((x + 3)^2 + (y – 4)^2 = r^2\)

📝 Example 1: Writing Equation from Center and Radius

Write the equation of a circle with center at \((2, -3)\) and radius 5.

Standard form: \((x – h)^2 + (y – k)^2 = r^2\)

Identify values: \(h = 2\), \(k = -3\), \(r = 5\)

Substitute: \((x – 2)^2 + (y – (-3))^2 = 5^2\)

Simplify: \((x – 2)^2 + (y + 3)^2 = 25\)

📝 Example 2: Finding Center and Radius

Find the center and radius of the circle: \((x + 4)^2 + (y – 1)^2 = 9\)

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

Identify h: \((x + 4)^2 = (x – (-4))^2\), so \(h = -4\)

Identify k: \((y – 1)^2\), so \(k = 1\)

Identify r: \(r^2 = 9\), so \(r = \sqrt{9} = 3\) (positive only)

Answer: Center = \((-4, 1)\), Radius = 3

Part 2: General Form and Conversion

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Standard Form ↔ General Form

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General Form of Circle Equation

The general form of a circle equation is:

\[x^2 + y^2 + 2gx + 2fy + c = 0\]

Where:

Center = \((-g, -f)\)
Radius = \(\sqrt{g^2 + f^2 – c}\)

Important: For this to represent a real circle, we need \(g^2 + f^2 – c > 0\) (positive radius squared).

Standard → General: Expand \((x-h)^2 + (y-k)^2 = r^2\)

Expand squares: \(x^2 – 2hx + h^2 + y^2 – 2ky + k^2 = r^2\)

Rearrange: \(x^2 + y^2 – 2hx – 2ky + (h^2 + k^2 – r^2) = 0\)

Compare with: \(x^2 + y^2 + 2gx + 2fy + c = 0\)
So \(2g = -2h \Rightarrow g = -h\), \(2f = -2k \Rightarrow f = -k\), \(c = h^2 + k^2 – r^2\)

📝 Example 3: Standard to General Form

Convert \((x – 3)^2 + (y + 2)^2 = 16\) to general form.

Expand: \((x^2 – 6x + 9) + (y^2 + 4y + 4) = 16\)

Combine: \(x^2 + y^2 – 6x + 4y + 13 = 16\)

Rearrange: \(x^2 + y^2 – 6x + 4y + 13 – 16 = 0\)

Final form: \(x^2 + y^2 – 6x + 4y – 3 = 0\)

Check: Compare with \(x^2 + y^2 + 2gx + 2fy + c = 0\)
Here: \(2g = -6 \Rightarrow g = -3\), \(2f = 4 \Rightarrow f = 2\), \(c = -3\)

📝 Example 4: General to Standard Form (Completing the Square)

Convert \(x^2 + y^2 – 8x + 6y + 16 = 0\) to standard form and find center and radius.

Group x and y terms: \((x^2 – 8x) + (y^2 + 6y) = -16\)

Complete square for x: \(x^2 – 8x = (x – 4)^2 – 16\)

Complete square for y: \(y^2 + 6y = (y + 3)^2 – 9\)

Substitute: \([(x – 4)^2 – 16] + [(y + 3)^2 – 9] = -16\)

Simplify: \((x – 4)^2 + (y + 3)^2 – 25 = -16\)

Rearrange: \((x – 4)^2 + (y + 3)^2 = 9\)

Identify: Center = \((4, -3)\), Radius = \(\sqrt{9} = 3\)

Part 3: Different Ways to Define a Circle

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Finding Equations from Different Information

Type 1: Center and Radius

Given: Center \((h, k)\), radius \(r\)

Equation: \((x – h)^2 + (y – k)^2 = r^2\)

Example: Center \((2, -1)\), radius 4
\((x – 2)^2 + (y + 1)^2 = 16\)

Type 2: Center and Point on Circle

Given: Center \((h, k)\), point \((x_1, y_1)\) on circle

Method: Calculate radius using distance formula:
\(r = \sqrt{(x_1 – h)^2 + (y_1 – k)^2}\)

Then use: \((x – h)^2 + (y – k)^2 = r^2\)

Type 3: Endpoints of Diameter

Given: Diameter endpoints \((x_1, y_1)\) and \((x_2, y_2)\)

Center: Midpoint of diameter
\(h = \frac{x_1 + x_2}{2}\), \(k = \frac{y_1 + y_2}{2}\)

Radius: Half the distance between endpoints

Type 4: Three Points on Circle

Given: Three points \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\)

Method: Substitute into general form
\(x^2 + y^2 + 2gx + 2fy + c = 0\)
Solve system of 3 equations for \(g, f, c\)

📝 Example 5: Center and Point on Circle

Find the equation of a circle with center at \((1, 4)\) that passes through the point \((-2, 6)\).

Find radius: Distance from center to point:
\(r = \sqrt{(-2 – 1)^2 + (6 – 4)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}\)

Write equation: \((x – 1)^2 + (y – 4)^2 = (\sqrt{13})^2\)

Simplify: \((x – 1)^2 + (y – 4)^2 = 13\)

📝 Example 6: Endpoints of Diameter (Past Paper Style)

A circle has diameter with endpoints \(A(1, 5)\) and \(B(7, 1)\). Find the equation of the circle.

Find center (midpoint):
\(h = \frac{1 + 7}{2} = 4\), \(k = \frac{5 + 1}{2} = 3\)
Center = \((4, 3)\)

Find radius (half of diameter length):
Diameter length = \(\sqrt{(7-1)^2 + (1-5)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}\)
Radius = \(\frac{2\sqrt{13}}{2} = \sqrt{13}\)

Write equation: \((x – 4)^2 + (y – 3)^2 = (\sqrt{13})^2\)

Simplify: \((x – 4)^2 + (y – 3)^2 = 13\)

Part 4: Circle Properties and Applications

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Working with Circle Equations

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Determining if a Point Lies on a Circle

To check if point \((x_1, y_1)\) lies on circle \((x – h)^2 + (y – k)^2 = r^2\):

Substitute \(x_1\) for \(x\) and \(y_1\) for \(y\)
Calculate \((x_1 – h)^2 + (y_1 – k)^2\)
If result equals \(r^2\), point lies on circle
If result < \(r^2\), point is inside circle
If result > \(r^2\), point is outside circle

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Finding Points on a Circle

Given \(x\)-coordinate, find \(y\)-coordinate(s) on circle \((x – h)^2 + (y – k)^2 = r^2\):

Substitute known \(x\) value into equation

Solve for \(y\): \((y – k)^2 = r^2 – (x – h)^2\)

Take square root: \(y – k = \pm \sqrt{r^2 – (x – h)^2}\)

Solve: \(y = k \pm \sqrt{r^2 – (x – h)^2}\)

Note: There are usually 2 \(y\)-values (except at top/bottom of circle)

📝 Example 7: Point Position Relative to Circle

Determine whether point \(P(2, 5)\) lies inside, on, or outside the circle \((x – 1)^2 + (y – 3)^2 = 9\).

Substitute: \((2 – 1)^2 + (5 – 3)^2 = 1^2 + 2^2 = 1 + 4 = 5\)

Compare with \(r^2\): \(r^2 = 9\)

Since 5 < 9: Point \(P(2, 5)\) is inside the circle

Real-World Applications:

Navigation: GPS systems use circle equations to determine locations within certain radii

Engineering: Designing circular components like gears, wheels, and pipes

Computer Graphics: Drawing circles and curves in software applications

Physics: Modeling circular motion and wave fronts

Architecture: Designing arches, domes, and circular structures

Part 5: Special Cases and Common Mistakes

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Avoiding Common Errors

Common Mistakes to Avoid: 1. Forgetting to square the radius in standard form
2. Incorrect signs when identifying center from equation
3. Taking negative square root for radius (radius is always positive)
4. Not completing the square properly when converting general to standard form
5. Confusing \(h\) and \(k\) signs: \((x – h)^2\) means h is positive, \((x + a)^2 = (x – (-a))^2\)
6. Forgetting that \(r^2\) must be positive in \((x-h)^2 + (y-k)^2 = r^2\)

Memory Aid for Standard Form: “MINUS, MINUS, PLUS”
\((x \ \text{MINUS} \ h)^2 + (y \ \text{MINUS} \ k)^2 = r^2\)
The center is \((h, k)\) with the signs as they appear in the parentheses after x and y.

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Recognizing Circle Equations

Not every equation of the form \(x^2 + y^2 + ax + by + c = 0\) represents a circle! Check if:

For \(x^2 + y^2 + 2gx + 2fy + c = 0\):
If \(g^2 + f^2 – c > 0\): Real circle
If \(g^2 + f^2 – c = 0\): Point circle (radius 0)
If \(g^2 + f^2 – c < 0\): No real circle (imaginary)

📝 Example 8: Does This Equation Represent a Circle?

Determine if \(x^2 + y^2 – 6x + 4y + 20 = 0\) represents a real circle.

Identify coefficients: Compare with \(x^2 + y^2 + 2gx + 2fy + c = 0\)
\(2g = -6 \Rightarrow g = -3\), \(2f = 4 \Rightarrow f = 2\), \(c = 20\)

Calculate: \(g^2 + f^2 – c = (-3)^2 + 2^2 – 20 = 9 + 4 – 20 = -7\)

Since -7 < 0: This does NOT represent a real circle (it’s an imaginary circle)

Comparison Table: Circle Equation Forms

Form	Equation	Center	Radius	When to Use
Standard Form	\((x – h)^2 + (y – k)^2 = r^2\)	\((h, k)\)	\(r\)	When center and radius are known or needed
General Form	\(x^2 + y^2 + 2gx + 2fy + c = 0\)	\((-g, -f)\)	\(\sqrt{g^2 + f^2 – c}\)	When given expanded form or need to find intercepts
Center at Origin	\(x^2 + y^2 = r^2\)	\((0, 0)\)	\(r\)	When circle is centered at origin
Diameter Form	\((x – x_1)(x – x_2) + (y – y_1)(y – y_2) = 0\)	Midpoint of \((x_1, y_1)\) and \((x_2, y_2)\)	Half the distance between points	When endpoints of diameter are given

Quiz: Test Your Understanding

Equations of a Circle Quiz

Question 1: Find the center and radius of the circle: \((x + 5)^2 + (y – 2)^2 = 36\)

Answer:
Compare with \((x – h)^2 + (y – k)^2 = r^2\)
\((x + 5)^2 = (x – (-5))^2\), so \(h = -5\)
\((y – 2)^2\), so \(k = 2\)
\(r^2 = 36 \Rightarrow r = 6\) (positive only)
Center = \((-5, 2)\), Radius = 6

Question 2: Write the equation of a circle with center at \((-3, 4)\) and radius 7.

Answer:
Using \((x – h)^2 + (y – k)^2 = r^2\)
Here \(h = -3\), \(k = 4\), \(r = 7\)
\((x – (-3))^2 + (y – 4)^2 = 7^2\)
\((x + 3)^2 + (y – 4)^2 = 49\)

Question 3: Convert to standard form and find center and radius: \(x^2 + y^2 – 10x + 6y + 30 = 0\)

Answer:
1. Group: \((x^2 – 10x) + (y^2 + 6y) = -30\)
2. Complete squares: \((x – 5)^2 – 25 + (y + 3)^2 – 9 = -30\)
3. Simplify: \((x – 5)^2 + (y + 3)^2 – 34 = -30\)
4. Final: \((x – 5)^2 + (y + 3)^2 = 4\)
Center = \((5, -3)\), Radius = 2

Question 4: A circle has diameter with endpoints \((2, 1)\) and \((8, 9)\). Find its equation.

Answer:
1. Center (midpoint): \(h = \frac{2+8}{2} = 5\), \(k = \frac{1+9}{2} = 5\)
2. Radius (half of diameter): Diameter = \(\sqrt{(8-2)^2 + (9-1)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)
Radius = 5
3. Equation: \((x – 5)^2 + (y – 5)^2 = 25\)

Question 5: Determine if point \((6, -2)\) lies inside, on, or outside the circle \((x – 4)^2 + (y + 3)^2 = 25\).

Answer:
Substitute: \((6 – 4)^2 + (-2 + 3)^2 = 2^2 + 1^2 = 4 + 1 = 5\)
Compare with \(r^2 = 25\)
Since \(5 < 25\), point \((6, -2)\) lies inside the circle.

🎯 Key Concepts Summary

Standard Form: \((x – h)^2 + (y – k)^2 = r^2\)
- Center = \((h, k)\)
- Radius = \(r\) (always positive)
- Remember: \((x + a)^2 = (x – (-a))^2\)
General Form: \(x^2 + y^2 + 2gx + 2fy + c = 0\)
- Center = \((-g, -f)\)
- Radius = \(\sqrt{g^2 + f^2 – c}\)
- Real circle only if \(g^2 + f^2 – c > 0\)
Conversion Methods:
- Standard → General: Expand and simplify
- General → Standard: Complete the square
Common CSEC Questions:
- Find center and radius given equation
- Write equation given center and radius
- Write equation given endpoints of diameter
- Convert between standard and general forms
- Determine if point lies on/inside/outside circle
Exam Strategy:
- Always show steps when converting forms
- Check your radius is positive
- When completing square: \(x^2 + bx = (x + \frac{b}{2})^2 – (\frac{b}{2})^2\)
- For diameter problems: center is midpoint, radius is half the distance

CSEC Exam Strategy: When answering circle equation questions: (1) Write the appropriate formula first, (2) Carefully identify h, k, and r (watch signs!), (3) When converting general to standard form, show completing the square steps clearly, (4) For diameter problems, find midpoint for center and half the distance for radius, (5) Always check that your radius squared is positive.