Chapter 05: Circles – Part 1: The Equation of a Circle, Centre and Radius

CHAPTER 05: CIRCLES

PART 1: THE EQUATION OF A CIRCLE, CENTRE AND RADIUS

A-Level Pure Mathematics | Mathematics HipHop Academy

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1. Learning Objectives

By the end of this lesson, you should be able to:

• Define a circle in terms of its centre and radius.
• Understand the standard equation of a circle.
• Identify the centre and radius from a circle equation.
• Write the equation of a circle when its centre and radius are given.
• Understand the general expanded form of a circle equation.
• Determine whether a given equation represents a real circle, a point or no real circle.

2. Introduction to Circles

A circle is the locus (path) of a point that moves in a plane while remaining at a constant distance from a fixed point.

The fixed point is called the centre of the circle, while the constant distance is called the radius.

Important terms

• Centre: The fixed point from which every point on the circle is equally distant.
• Radius: The distance from the centre to any point on the circumference.
• Diameter: A straight line through the centre joining two points on the circumference. Its length is twice the radius.
• Circumference: The boundary of the circle.

If the radius is \(r\), then the diameter is \(2r\).

3. The Distance Formula

Before developing the equation of a circle, recall the distance formula between two points.

If \(A(x_1,y_1)\) and \(B(x_2,y_2)\), then:

\(AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

This formula follows from Pythagoras' theorem and is essential when finding the equation of a circle.

4. Deriving the Standard Equation of a Circle

Consider a circle with centre \(C(a,b)\) and radius \(r\).

Let \(P(x,y)\) be any point on its circumference.

Since every point on the circumference is a distance \(r\) from the centre:

\(CP=r\)

Using the distance formula:

\(\sqrt{(x-a)^2+(y-b)^2}=r\)

Squaring both sides gives:

STANDARD EQUATION OF A CIRCLE

\((x-a)^2+(y-b)^2=r^2\)

Centre: \((a,b)\) | Radius: \(r\)

This is the most important equation in this part of the chapter.

5. Circle with Centre at the Origin

If the centre is at the origin \(O(0,0)\), then \(a=0\) and \(b=0\).

Substituting into the standard equation:

\((x-0)^2+(y-0)^2=r^2\)

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

This represents a circle centred at the origin with radius \(r\).

6. Identifying the Centre and Radius

When an equation is written in the form:

\((x-a)^2+(y-b)^2=r^2\)

we can read the centre and radius directly:

• Centre = \((a,b)\)
• Radius = \(r\)

Be careful: The signs inside the brackets are opposite to the coordinates of the centre.

For example, \((x-3)^2+(y+4)^2=25\) has centre \((3,-4)\), not \((3,4)\).

7. Worked Examples

Example 1: Identifying the Centre and Radius

Find the centre and radius of the circle:

\((x-4)^2+(y-2)^2=36\)

Solution

Compare with the standard equation:

\((x-a)^2+(y-b)^2=r^2\)

Therefore:

• \(a=4\)
• \(b=2\)
• \(r^2=36\), so \(r=6\)

Answer: Centre = (4, 2), Radius = 6 units.

Example 2: Negative Coordinates of the Centre

Find the centre and radius of:

\((x+5)^2+(y-3)^2=49\)

Solution

Rewrite mentally in standard form:

\((x-(-5))^2+(y-3)^2=7^2\)

Therefore:

• Centre = \((-5,3)\)
• Radius = \(7\)

Answer: Centre = (-5, 3), Radius = 7 units.

Example 3: Forming an Equation from Centre and Radius

Find the equation of a circle whose centre is \((2,-3)\) and whose radius is \(5\).

Solution

Use:

\((x-a)^2+(y-b)^2=r^2\)

Substitute \(a=2\), \(b=-3\) and \(r=5\):

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

Answer: \((x-2)^2+(y+3)^2=25\)

Example 4: Finding the Equation from a Point on the Circle

A circle has centre \(C(1,2)\) and passes through \(P(4,6)\). Find its equation.

Solution

First calculate the radius using the distance formula:

\(r=\sqrt{(4-1)^2+(6-2)^2}\)

\(r=\sqrt{9+16}=\sqrt{25}=5\)

Substitute the centre and radius into the standard equation:

\((x-1)^2+(y-2)^2=5^2\)

Answer: \((x-1)^2+(y-2)^2=25\)

Example 5: Checking Whether a Point Lies on a Circle

Determine whether the point \(P(3,4)\) lies on the circle:

\(x^2+y^2=25\)

Solution

Substitute \(x=3\) and \(y=4\):

\(3^2+4^2=9+16=25\)

The equation is satisfied. Therefore, the point lies on the circumference.

Answer: Yes, P(3, 4) lies on the circle.

8. The General Expanded Form of a Circle

The standard equation can be expanded to obtain another useful form.

Starting with:

\((x-a)^2+(y-b)^2=r^2\)

Expand the brackets:

\(x^2-2ax+a^2+y^2-2by+b^2=r^2\)

Collect the terms:

\(x^2+y^2-2ax-2by+a^2+b^2-r^2=0\)

Let:

\(g=-a,\quad f=-b,\quad c=a^2+b^2-r^2\)

GENERAL FORM

\(x^2+y^2+2gx+2fy+c=0\)

For this form:

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

The expression under the square root determines the nature of the circle.

9. Nature of the Circle

For the general equation:

\(x^2+y^2+2gx+2fy+c=0\)

the radius satisfies:

\(r^2=g^2+f^2-c\)

• If \(g^2+f^2-c>0\), there is a real circle with positive radius.
• If \(g^2+f^2-c=0\), the radius is zero and the circle reduces to a single point.
• If \(g^2+f^2-c<0\), there is no real circle.

10. Worked Examples on the General Form

Example 6: Finding the Centre and Radius from the Expanded Form

Find the centre and radius of:

\(x^2+y^2-6x+8y-11=0\)

Solution

Compare with:

\(x^2+y^2+2gx+2fy+c=0\)

We have:

• \(2g=-6\), so \(g=-3\)
• \(2f=8\), so \(f=4\)
• \(c=-11\)

Centre = \((-g,-f)=(3,-4)\).

Radius:

\(r=\sqrt{g^2+f^2-c}\)

\(=\sqrt{(-3)^2+4^2-(-11)}\)

\(=\sqrt{9+16+11}=\sqrt{36}=6\)

Answer: Centre = (3, -4), Radius = 6 units.

Example 7: Determining Whether an Equation Represents a Real Circle

Determine the nature of:

\(x^2+y^2+4x-6y+20=0\)

Solution

Comparing with the general form:

\(2g=4\Rightarrow g=2\)

\(2f=-6\Rightarrow f=-3\)

\(c=20\)

Calculate \(r^2\):

\(r^2=g^2+f^2-c\)

\(=2^2+(-3)^2-20\)

\(=4+9-20=-7\)

Since \(r^2\) is negative, no real radius exists.

Answer: The equation does not represent a real circle.

11. Common Errors to Avoid

• Incorrect centre signs: The centre of \((x+2)^2+(y-5)^2=16\) is \((-2,5)\), not \((2,-5)\).
• Confusing radius and radius squared: If \(r^2=49\), then \(r=7\), not 49.
• Forgetting the square root: When calculating the radius from the general form, take the square root of \(g^2+f^2-c\).
• Ignoring the nature of the radius: A negative value of \(r^2\) does not give a real circle.
• Not showing working: In examinations, show the comparison of coefficients and the calculation of the radius.

12. Practice Exercises

Attempt all the questions before opening the answers. Show your working clearly.

Exercise A: Basic Understanding

1. Find the centre and radius of \((x-2)^2+(y-5)^2=64\).
2. Find the centre and radius of \((x+4)^2+(y+3)^2=81\).
3. Write the equation of a circle with centre \((3,-2)\) and radius \(4\).
4. Write the equation of a circle centred at the origin with radius \(9\).
5. Find the radius of a circle whose diameter is \(18\) units.

Exercise B: Application

6. A circle has centre \((2,1)\) and passes through \((5,5)\). Find its equation.
7. Determine whether the point \((2,3)\) lies on the circle \(x^2+y^2=13\).
8. Find the centre and radius of \(x^2+y^2-10x+4y-20=0\).
9. Find the centre and radius of \(x^2+y^2+8x-2y+1=0\).
10. Determine whether \(x^2+y^2-2x+4y+10=0\) represents a real circle.

Exercise C: Examination Challenge

11. Find the equation of the circle whose centre is \((-3,2)\) and which passes through the point \((1,5)\).
12. A circle has equation \(x^2+y^2+2gx+2fy+13=0\). If its centre is \((2,-3)\), find \(g\), \(f\) and its radius.
13. Find the values of \(k\) for which \(x^2+y^2-4x+6y+k=0\) represents:
    1. a real circle with positive radius;
    2. a circle reduced to a point;
    3. no real circle.

OPEN ANSWERS – EXERCISE A

1. Centre \((2,5)\), radius \(8\).
2. Centre \((-4,-3)\), radius \(9\).
3. \((x-3)^2+(y+2)^2=16\).
4. \(x^2+y^2=81\).
5. Radius \(=18/2=9\) units.

OPEN ANSWERS – EXERCISE B

6. Radius \(=\sqrt{(5-2)^2+(5-1)^2}=5\). Equation: \((x-2)^2+(y-1)^2=25\).
7. Yes, because \(2^2+3^2=13\).
8. Centre \((5,-2)\), radius \(7\).
9. Centre \((-4,1)\), radius \(4\).
10. No real circle, because \(r^2=1+4-10=-5\).

OPEN ANSWERS – EXERCISE C

11. Radius \(=\sqrt{(1+3)^2+(5-2)^2}=5\). Equation: \((x+3)^2+(y-2)^2=25\).
12. Centre \((-g,-f)=(2,-3)\), so \(g=-2\), \(f=3\). Then \(r^2=4+9-13=0\). Radius \(=0\); it is a point circle.
13. Here \(g=-2\), \(f=3\). Thus \(r^2=4+9-k=13-k\).
    1. Positive radius: \(k<13\).
    2. Point circle: \(k=13\).
    3. No real circle: \(k>13\).

13. Part 1 Summary

Standard equation: \((x-a)^2+(y-b)^2=r^2\)

Centre: \((a,b)\)

Radius: \(r\)

Circle at origin: \(x^2+y^2=r^2\)

General form: \(x^2+y^2+2gx+2fy+c=0\)

Centre in general form: \((-g,-f)\)

Radius in general form: \(\sqrt{g^2+f^2-c}\)

Coming Next: Part 2

In Part 2, we shall study Completing the Square and Finding Circle Equations. We will learn how to transform expanded equations into standard form and use given conditions to determine unknown circle equations.

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