# 3. The Circle

General form of a Circle

## a. Circle Formulas

### Center at the Origin

The circle with centre (0, 0) and radius r has the equation:

x2 + y2 = r2

This means any point (x, y) on the circle will give the radius squared when substituted into the circle equation.

### Center not at the Origin

The circle with centre (h, k) and radius r has the equation:

(xh)2 + (yk)2 = r2

These formulas are a direct result of Pythagoras' Formula for the length of the hypotenuse of a right triangle.

### Example 1

Sketch the circle x2 + y2 = 4.

Find the center and radius first.

The equation is in the form x2 + y2 = r2, so we have a circle, center (0, 0) with radius r=sqrt4=2.

### Example 2

Sketch the circle (x − 2)2 + (y − 3)2 = 16

Find the center and radius first.

The equation is in the form (xh)2 + (yk)2 = r2, so we have a circle with centre at (2, 3) and the radius is r=sqrt(16)=4.

### Example 3

Sketch the circle (x + 4)2 + (y − 5)2 = 36

The equation is in the form (xh)2 + (yk)2 = r2.

We have a circle with centre at (−4, 5) and the radius is r=sqrt(36)=6.

Be careful with positive and negative x- and y-values in this work!

## b. The General Form of the Circle

An equation which can be written in the following form (with constants D, E, F) represents a circle:

x2 + y2 + Dx + Ey + F = 0

This is called the general form of the circle.

### Example 4

Find the centre and radius of the circle

x2 + y2 + 8x + 6y = 0

Sketch the circle.

Please revise Completing the Square first...

Our aim is to get the equation into the form: (xh)2 + (yk)2 = r2

We complete the square on the x-related portion and on the y-related portion, at the same time.

x^2+y^2+8x+6y=0

Group the x parts together and the y parts togther:

(x^2+8x)+(y^2+6y)=0

Complete the square on each of the x and y parts.

(x^2+8x+16)+(y^2+6y+9) =16+9 =25

(x+4)^2+(y+3)^2=5^2

This is now in the format we require and we can determine the center and radius of the circle.

So the centre of the circle is (−4, −3) and the radius is 5 units.

Note that the circle passes through (0, 0). This is logical, since:

1. The circle has radius 5
2. Considering the right triangle formed by the points (−4, −3), (0, −3), and (0, 0), we can apply Pythagoras' Theorem and obtain: (−4)2 + (−3)2 = (5)2
3. The formula for a circle follows from Pythagoras' Theorem.

### Exercises

1. Find the equation of the circle with centre (3/2, -2) and radius 5/2.

Centre (3/2, -2) radius 5/2.

General form of the equation of a circle:

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

The required equation for this case:

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

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

There is no need to expand this out, since this is the most useful form of the equation.

2. Determine the centre and radius and then sketch the circle:

3x2 + 3y2 − 12x + 4 = 0

We complete the square as we did in an earlier example above.

First, we collect the x parts together and the y parts together, then divide throughout by 3.

3x^2+3y^2-12x+4=0

3x^2-12x+3y^2+4=0

x^2-4x+y^2+4/3=0

Then we complete the square on the x part. We do not need to do so for the y part because there is no single y term (only a y2 term).

(x^2-4x+4)+y^2+4/3=4

(x-2)^2+y^2=4-4/3

(x-2)^2+y^2=8/3

So the circle has centre (2,0) and has radius sqrt(8/3)~~1.63.

3. Find the points of intersection of the circle

x2 + y2x − 3y = 0

with the line

y = x − 1.

We solve the 2 equations simultaneously by substituting the expression

y = x -1

into the expression

x^2+y^2-x-3y=0

[See some background to this at: Algebraic Solution of Systems of Equations.]

We have:

x^2+(x-1)^2-x-3(x-1)=0

x^2+x^2-2x+1-x-3x+3=0

2x^2-6x+4=0

x^2-3x+2=0

(x-1)(x-2)=0

So we see that the solutions for x are x = 1 or x = 2. This gives the corresponding y-values of y = 0 and y = 1.

So the points of intersection are at: (1, 0) and (2, 1).

We can see that our answer is correct in a sketch of the situation:

### Exercise for You

Where did (0.5, 1.5) for the center of the circle come from?

### Conic section: Circle

How can we obtain a circle from slicing a cone?

Each of the lines and curves in this chapter are conic sections, which means the curves are formed when we slice a cone at a certain angle. If we slice a cone with a plane at right angles to the axis of the cone, the shape formed is a circle.

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