COORDINATE GEOMETRY OF THE CIRCLE
1. The Centre of a Circle
In the Coordinate Geometry topic in unit C1, we said…
In mathematics, a line is infinitely long. A finite portion of a line is called a line segment.
It is obvious that the midpoint of the line segment joining the points and is
We now apply this to the circle.
Example1: The line AB is the diameter of a circle, where A and B are (3, 7) and (5, −4) respectively. Find the coordinates of the centre of the circle.
The centre of the circle is clearly the midpoint of the diameter.
Example 2:The centre of a circle C is (2, −7). The point A on the circumference of the circle is (5, 2). Find the coordinates of B, where AB is a diameter of the circle.
There are two ways of approaching this. We can use our formula for the midpoint of a line segment…
…and so B is (−1, −16).
Alternatively, we can realise that the journey from A to C is the same as the journey from C to B. Using vectors,
Therefore B has coordinates
C2 p51 Ex 4A
2. Chords
A chord is a line segment joining two points on the circumference of a circle. A diameter is the special case where the chord passes through the centre of the circle.
Theperpendicular from the centre of a circle to a chord bisects the chord. We say the line is the perpendicular bisector of the chord.
Example1:The lines AB and CD are chords of a circle. The line is the perpendicular bisector of AB. The line is the perpendicular bisector of CD. Find the coordinates of the centre of the circle.
Both perpendicular bisectors pass through the centre of the circle, so all we have to do is find where they meet.
The centre of the circle is (2, 3).
Example2:The centre of a circle is (−2, −5.5). Show that the line AB is a chord of the circle, where A is (5, 6) and B is (3, 7).
The perpendicular bisector of the line joining A and B is perpendicular to AB and passes through the midpoint of AB.The midpoint of AB is
The gradient of AB is . Therefore the gradient of the perpendicular bisector is 2.
To find the equation of the perpendicular bisector,
Substituting gives us , and therefore the centre of the circle lies on the perpendicular bisector of AB. This means that AB must be a chord to the circle.
C2 p56 Ex 4B
3. The Angle in a Semicircle
Remember from GCSE that the angle in a semicircle is a right angle.
This can be proved using isosceles triangles.
Using the large triangle, we have
But since is the angle in the semicircle, our theorem is proved.
Example 1 : The points A (–3, 1), B (1, 3) and C (4,–3) lie on a circle.
a) Prove that .
b) Find the coordinates of the centre of the circle.
a) We begin by finding the gradients of the lines AB and BC.
The product of the gradients is –1, and so the lines meet at a right angle.
b) We now know that AC is a diameter of the circle, so the centre of the circle is the midpoint of AC, namely .
C2 p59 Ex 4C
4. Cartesian Equation of a Circle
Consider a circle of radius r and centre (0, 0), and a point P on its circumference.
Since any point P, must be a distance r from the centre of the circle, we can use Pythagoras to obtain the Cartesian equation of the circle.
Now consider a circle of radius r and centre (a, b).
Pythagoras gives us the general Cartesian equation of the circle.
Example 1 : Find the centre and radius of the circle with equation
The centre is (3, –2) and the radius is 4.
Activity1:Verify this by using Autograph. The function may be entered in its implicit form. Click on the equal aspect icon so it looks like a circle and not like an ellipse!
All of the following examples can also be checked on Autograph, and its use is strongly recommended.
Example 2 : Find the centre and radius of the circle with equation
We complete the square to get the equation in the form.
The centre is (–1, 3) and the radius is 4.
Example3:Find where the circle with centre (–2, –5) and radius 7 units crosses the x-axis.
The equation of the circle is
The circle crosses the x-axis when .
The circle crosses the x-axis at.
Example 4 : Find the equation of the tangent to circle with equation at the point (3, 2).
The equation of the circle can be rearranged as follows
The centre of the circle is therefore (–3, 1). The gradient of the radius joining the centre to the point (3, 2) is
Since the tangent is perpendicular to the radius, the tangent has gradient –6. The equation of the tangent is of the form
Example5:Find the points of intersection (if any) of the circle and the line.
Substituting the equation of the line into the equation of the circle,
There is only one solution, (−1, −5), and so the line is a tangent to the circle at this point.
Example6:Find the points of intersection (if any) of the the circle and the line .
Substituting the equation of the line into the equation of the circle,
There are no solutions and therefore the line and the circle do not intersect.
Example7:Prove that part of the line forms a chord to the circle
and find the length of the chord.
Substituting the equation of the line into the equation of the circle,
The line intersects the circle at the points, and is therefore a chord to the circle. The two points are (0, 5) and (−1, 2). Using Pythagoras, the length of the chord is units.
Example 8 : A is (7, −2) and B is (−3, 10). Find the equation of the circle whose diameter is AB.
The centre of the circle is the midpoint of AB.
The radius rof the circle is the distance from the centre to A (or B), which by Pythagoras is given by
Therefore the equation of the circle is
Example9:Find the greatest and least distances from the point (1, 6) to the circle
We first find the centre and radius of the circle.
The circle has centre (5, 3) and radius 2. At this stage a diagram is very useful!
By Pythagoras, the distance from (1, 6) to the centre is . Therefore, the nearest point on the circle to (1, 6) is units away, and the furthest point is units away.
Example10:Find the equation, centre and radius of the circle passing through (0, 1), (4, 3) and (1, −1).
There are two methods here.
Brute force algebra method…
There are three unknowns in the equation of the circle (namely a, b and r), and so we need three simultaneous equations to find them.
Substituting each pair of coordinates intogives us
Equating the first and second equations gives us
Equating the first and third equations gives us
Solving simultaneously by subtracting the two equations in a and b,
Substituting these values into the first equation,
So the equation of the circle is
and the circle has centre (2.5, 1) and radius 2.5 units.
Perpendicular bisector method…
The centre of the circle is equidistant from all three points given, and hence will lie on all three perpendicular bisectors, although of course we need only find the equations of two of these bisectors.
We will first find the equation of the perpendicular bisector of (0, 1) and (4, 3).
The midpoint of these two points is (2, 2), and the gradient of the line between them is . The gradient of the perpendicular bisector is therefore −2, and the equation is
Next, we will find the equation of the perpendicular bisector of (0, 1) and (1, −1).The midpoint of these two points is (, 0), and the gradient of the line between them is . The gradient of the perpendicular bisector is therefore , and the equation is
The bisectors meet when
So the centre of the circle is (2.5, 1). To find the radius, we use Pythagoras to find the distance from the centre of the circle to any of the points, say (4, 3)
Finally, the equation of the circle is
Example11*:The line is a tangent to the circle. Find the two possible values of m.
We first substitute the equation of the straight line into the equation of the circle to find the points of intersection.
Since the line is a tangent, this quadratic equation can have only one solution, and therefore its discriminant is equal to zero.
C2p63 Ex 4D, p66 Ex 4E Topic Review : Coordinate Geometry of the Circle