The Equation Of A Circle
The Equation of a Circle – Introduction
AS Maths and circle geometry require a solid understanding of straight line geometry, algebra skills like simultaneous equations, and Pythagoras. These concepts are essential for solving problems related to circle geometry.
Straight line geometry plays a crucial role in circle geometry as circles can be defined and understood in relation to lines. Understanding the properties and equations of lines is necessary for working with circles effectively.
Algebra skills, specifically simultaneous equations, are important for solving problems involving circle geometry. Simultaneous equations can help find the intersection points between circles and lines, which are crucial in solving various circle-related problems.
Pythagoras’ theorem is another fundamental concept needed in circle geometry. It allows us to calculate the lengths of sides in right-angled triangles, which frequently arise in circle-related problems.
Overall, a solid grasp of straight line geometry, algebra skills including simultaneous equations, and Pythagoras is essential for tackling AS Maths and circle geometry effectively. These concepts provide the foundational knowledge required to solve problems and understand the properties and relationships within circle geometry.
The equation of a circle takes two forms and you need to be aware of both types.
The first in when the circle has a centre at the origin and radius, r. Then the equation of the circle takes the following form:
x^2+y^2=r^2The second form of the equation of a circle is where the centre is at a general point (a, b) and radius, r. Then the equation of the circle takes the following form:
(x-a)^2+(y-b)^2=r^2The Equation Of A Circle - 3 Theorems
In order to be successful with circle geometry questions there are 3 three theorems that you need to be aware of. These are not the same as the Circle Theorems topic that was covered at GCSE Level.
Theorem 1:
If you have a triangle in a semicircle, then the angle at C is a right angle and AB is the diameter.
Theorem 2:
The line from the centre bisects a chord at its midpoint.
Theorem 3:
A tangent to a circle is perpendicular to the line of the radius.
The above three theorems need to be kept in mind when doing any questions involving circle geometry.
The Equation Of A Circle - Easy Examples
Part a)
x^2+y^2=49Part b)
(x+2)^2+(y-6)^2=9For part a) because there are no brackets involved that you know that the centre is the (0, 0). To determine the radius simply find the square root of 49 which is 7.
For part b) in order to determine the coordinates of the centre of the circle simply swap the signs of the numbers. You will see a “+2” and “-6”. What is needed is for these signs to be reversed. So the centre will be (-2, 6). The radius of the circle will be the square root of 9 which is 3.
You can see that by completing the square, the standard equation of a circle has been obtained. So it has been shown that the original equation does indeed represent a circle.
The centre is therefore (-2, 3) and the radius is 4.
Finding the equation of a circle
Let us consider the following examples.
For part a) because we have a centre at the origin there will be no equation with brackets. The radius has been given to us with a value of 4 and therefore the equation of this circle will be as follows:
x^2+y^2=16For part b) the centre has been given. To get the brackets correct the signs need to be reversed and the radius needs to be squared. This will then give the following equation of a circle:
(x-3)^2+(y+4)^2=36The equation of a circle – slightly harder examples
When it comes to many questions in maths, especially that of geometry it is very important to be able to draw a quick diagram so as to understand and visualise the problem at hand.
Here is a quick sketch:
You can see the centre and you can see the point that is also on the circle. With questions like this you need to ask what other information is needed in order to answer the question? Well the piece of information that is missing is that of the radius and that can be calculated using Pythagoras’ Theorem as follows:
\begin{aligned} r^2 & =(1-(-2))^2+(-2-(-3))^2 \\ & =3^2+1^2 \\ & =10 \end{aligned}Which means that the radius is \sqrt{10}
This then gives the final answer as follows;
(x-1)^2+(y+2)^2=10The Equation of a Circle – Finding A Tangent
First you should sketch a diagram such as the one shown:
You can quickly determine the centre of the circle and from the diagram you can see that the tangent is perpendicular to the line of the tangent. The first thing that you can do is to find the gradient of the radius line which is as follows:
Gradient =\frac{1-(-2)}{4-3}=\frac{3}{1}=3
This then means that the gradient of the tangent is the negative reciprocal so –13.
You are told that the tangent passes through the point (4, 1) and so the equation of the tangent can be calculated as follows:
\begin{aligned} & y-1=-\frac{1}{3}(x-4) \\ & 3 y-3=-x+4 \\ & 3 y+x=7 \end{aligned}The Equation Of A Circle – Intersections
In order to find where a straight line intersects a circle or where a circle intersects another circle then you need to do simultaneous equations. In either case you need to use the method of substitution where you will obtain a quadratic.
You need to be very careful with the algebra here as it can be very easy to make a mistake and if dealing with a straight line, try to avoid a rearrangement that will involve the use of fractions, if possible. Let us consider a few examples before looking at some exam style questions.
This quite a straightforward example as both equations are equal to “y” so it makes to simply equate them as follows and to solve the quadratic:
\begin{aligned} & x^2-3 x+5=x+2 \\ & x^2-4 x+3=0 \\ & (x-3)(x-1)=0 \\ & x=3 \text { or } x=1 \end{aligned}Once you have determined the “x” values remember to find the “y” values by performing a simple substitution as shown:
\begin{aligned} & x=3 \text { then } y=3+2=5 \\ & x=1 \text { then } y=1+2=3 . \end{aligned}The coordinates are therefore (3,5) \text { and }(1,3)
Again with these two equations, as they are both equal to “y” the simplest approach is to equate them and to solve the quadratic as shown:
\begin{aligned} & x^2-6 x+5=-2 x^2+12 x-19 \\ & 3 x^2-18 x+24=0 \\ & x^2-6 x+8=0 \\ & (x-2)(x-4)=0 \\ & x=2 \text { or } x=4 \end{aligned}And the “y” values can then also be obtained as follows:
\begin{aligned} & x=2, y=2^2-6 \times 2+5=-3 \\ & x=4, y=4^2-6 \times 4+5=-3 \end{aligned}The points of intersection are (2,-3) \text { and }(4,-3)
The Equation of a Circle – Exam Style Questions
For part a) here is a quick sketch of the circle and the information that is given:
In order to find the equation the piece of information that is missing is that of the radius and that can be found through Pythagoras and this is done as follows:
\begin{aligned} & r^2=3^2+5^2 \\ & r^2=34 \\ & r=\sqrt{34} \end{aligned}So the equation of the circle can be written as:
(x+2)^2+(y-3)^2=34Part b)
To show that (3, 6) lies on the circle, simply substitute these values into the equation of the circle obtained. Both the left and the right should then equal the same number.
\begin{aligned} (3+2)^2+(6-3)^2 & =34 \\ 34 & =34 \end{aligned}Part c)
Here is a new scenario in that we are asked to determine the equation of a tangent to the circle. Drawing a sketch this would like the following:
Because the tangent is perpendicular to the radius of the line, the gradient of the line needs to be found. From that you can determine the gradient of the tangent using the negative reciprocal and then find the equation of the tangent. Note how the question has asked for the gradient of the tangent to be given and it must be in the form that the question has asked.
Gradient of radius line is calculated as:
\begin{aligned} & =\frac{6-3}{3–2} \\ & =\frac{3}{5} \end{aligned}The gradient of the tangent and its equation can be found as follows:
\begin{aligned} & y=-\frac{5}{3} x+c \\ & 6=-\frac{5}{3}(3)+c \\ & 6=-5+c \\ & c=11 \\ & y=-\frac{5}{3} x+11 \\ & 3 y=-5 x+33 \\ & 5 x+3 y-33=0 \end{aligned}With circle geometry questions it is always important to be drawing sketches and to always have at the front of your mind the three theorem’s that were mentioned. You will also still need to be able to use your straight line geometry skills when answering these questions.
If you are finding A Level Maths a challenge then our online tutors for maths are able to help. It is important to remember that geometry is a very visual subject so do questions with a diagram. You can save yourself a great deal of time.