# Circles (Additional Math)

Updated: Jun 22, 2021

When we are dealing with Additional Mathematics questions on circles, we must be able to identify the different forms of an equation of a circle, and know its different components which can yield us certain useful information and characteristic of the circle to help us with solving the questions.

There are generally two forms that an equation of a circle takes: **Standard Form** and **General Form**.

__In this note, you will learn:__

**- Different forms of an equation of a circle**

**1) Standard Form**

**2) General Form**

__1) Standard Form__

__1) Standard Form__

The standard form of an equation for a circle is given as such:

**(x – a)^2** **+ (y – b)^2** **= r^2, where r > 0, (a, b) is the center and r is the radius**

How we derive this equation is by making use of two points on the circle itself, and forming an equation to calculate the distance between the two points - one of it is the center of the circle and the other is on the circumference of the circle, which the distance will be equivalent to the radius of the circle, **r**.

Let’s take a look at how it’s done!

Given that we have two points on the circle **(x, y) – point on the circumference of the circle** and **(a, b) – point on the center of the circle.**

To find the distance between two points on a graph, the formula is given by:

**d =** **√ [(x2** **– x1)^2** **+ (y2** **– y1)^2]**

Hence, the radius of the circle will hence be: **r =** **√ [(x – a)^2** **+ (y – b)^2]**

We will square both sides of the equation to get rid of the square root and we will end up with:

**r^2** **=** **(x – a)^2** **+ (y – b)^2**

And that is how we derive our equation of circle in its standard form:

**(x – a)^2** **+ (y – b)^2** **=** **r^2**

__2) General Form__

__2) General Form__

Students are usually capable of identifying the center of a circle and its radius easily when given the equation of a circle in its standard form.

However, questions will typically ask students to find the center of the circle and its radius when the equation of the circle is given in its **general form**, which is given by:

**x^2** **+ y^2** **+ 2gx + 2fy + c = 0**

**where g^2** **+ f^2** **– c > 0, (-g, -f) is the center and √ [g2** **+ f2** **– c] is the radius**

When we are given an equation of a circle in its general form, it is impossible to identify its center and radius just by looking at the equation, hence we must convert it back into its **standard form** first by using **completing the square** method. Let’s see how it’s done!

Given that we have an equation of a circle in its general form,

x^2 + y^2 + 2gx + 2fy + c = 0

Firstly, let’s group the terms with the same variables together and shift the constant term to the other side:

(x^2 + 2gx) + (y^2 + 2fy) = -c

Then, by completing the square, we will get:

(x^2 + 2gx + **g^2**) + (y^2 + 2fy + **f^2**) = -c + **g^2** + **f^2**

(Note: we added the terms ‘**g^2**’ and ‘**f^2**’ to the equation because we want to get the quadratic form of **a^2** **+ 2ab + b^2** **= (a + b)^2** to resemble the standard form of a circle, and we must remember that whatever we add on the left side, we must add it to the right side as well!)

We can simplify the above equation as so:

(x + g)^2 + (y + f)^2 = g^2 + f^2 – c

Now, we can compare the equation of a circle in its standard form and in its general form to find the center of the circle and its radius!

**(x – a)^2** **+ (y – b)^2** **=** **r^2**

**(x + g)^2** **+ (y + f)^2** **= g^2** **+ f^2** **– c**

From the two equations above, we can see that **a = -g**, **b = -f** and **r = √ [g2** **+ f2** **– c].**

And this is why **(-g, -f)** is the center of the circle and **√ [g2** **+ f2** **– c]** is its radius!

Let’s take a look at how this works with an example:

Given that we have an equation **x^2** **+ y^2** **– 2x – 4y – 4 = 0**, find the coordinates of the center of the circle and its radius.

We can clearly see that this equation is in its **general form**, and hence we must use the **completing the square** method to change it to standard form in order for us to obtain the coordinates of the center of the circle and its radius.

**x^2** **+ y^2** **– 2x – 4y – 4 = 0**

**We will group the terms that have the same variables together and shift the constant to the other side:**

**(x^2** **- 2x) + (y^2** **– 4y) = 4**

**Now, by completing the square:**

**[(x^2** **- 2x + (-1)^2)] + [(y^2** **– 4y + (-2)^2)] = 4 + (-1)^2** **+ (-2)^2**

**(x^2** **– 2x + 1) + (y^2** **– 4y + 4) = 9**

**(x – 1)^2** **+ (y – 2)^2** **= 9**

**From the equation, you can see that a = 1, b = 2 and r = √ 9 = 3**

**Therefore, the coordinates of the center of the circle is (1 ,2) and its radius is 3 units.**

__Question Time!__

**1.** The equation of a circle is given by **(x + 5)^2** **+ (y – 7)^2** **= 144**, write down the coordinates of the center of the circle and its radius.

**2.** Given that the center of a circle is (5, 8) and has a radius of 3 cm, express the equation of the circle in its standard and general form.

**3.** Find the equation of the circle which passes through the points **A (4, 0)** and **B (5, 8)**, and has its center lying on the line **4y + 2x = 8**.

**4.** The line **2y + x = 5** intersects a circle with the equation of **x^2** **+ y^2** **– 4x – 16y – 32 = 0**, find the coordinates of the points of intersection.

And that’s all for today, students! ** Math Lobby** hopes that after this article, you have a clear understanding on the topic of circles in additional mathematics, and is equipped with the necessary skills to deal with questions involving finding the center and radius of a circle when given in both the standard and general forms!

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