Welcome to the World of Circles!
In this chapter, we are moving from straight lines to curves—specifically, the most perfect shape in nature: the circle. While you have been drawing circles with a compass for years, we are now going to describe them using Coordinate Geometry. By the end of this section, you will be able to look at an equation and "see" exactly where a circle sits on a graph and how big it is.
Don't worry if this seems like a big jump from straight lines. If you can use Pythagoras' Theorem and solve basic equations, you already have the tools you need to master this!
1. The Equation of a Circle
The standard way to write the equation of a circle with centre \( (a, b) \) and radius \( r \) is:
\( (x - a)^2 + (y - b)^2 = r^2 \)
Why does it look like this?
Think of a circle as a collection of points that are all the same distance (\( r \)) from a fixed centre point. If you draw a right-angled triangle from the centre to any point on the edge, Pythagoras' Theorem (\( A^2 + B^2 = C^2 \)) gives us this formula!
How to use the formula:
- If the centre is \( (3, 4) \) and the radius is \( 5 \), the equation is: \( (x - 3)^2 + (y - 4)^2 = 25 \).
- Watch the signs! If the equation is \( (x + 2)^2 + (y - 1)^2 = 16 \), the centre is actually \( (-2, 1) \) because the formula uses "minus \( a \)".
- The Radius: The number on the right side of the equals sign is \( r^2 \). To find the actual radius, you must take the square root. In the example above, the radius is \( \sqrt{16} = 4 \).
Quick Review:
Equation: \( (x - a)^2 + (y - b)^2 = r^2 \)
Centre: \( (a, b) \)
Radius: \( r \)
2. Expanding and Completing the Square
Sometimes, exam questions won't give you the nice, neat brackets. Instead, they might give you something like this:
\( x^2 + y^2 - 6x + 8y + 9 = 0 \)
To find the centre and radius from this "messy" version, we use a technique called Completing the Square. We do this for the \( x \) parts and the \( y \) parts separately.
Step-by-Step Guide:
Example: Find the centre and radius of \( x^2 + y^2 - 6x + 8y + 9 = 0 \).
- Group the terms: Put the \( x \)'s together and the \( y \)'s together. Move the plain number to the other side.
\( (x^2 - 6x) + (y^2 + 8y) = -9 \) - Complete the square for \( x \): Half of \(-6\) is \(-3\). So, we write \( (x - 3)^2 \), then subtract \((-3)^2\).
\( (x - 3)^2 - 9 \) - Complete the square for \( y \): Half of \( 8 \) is \( 4 \). So, we write \( (y + 4)^2 \), then subtract \( (4)^2 \).
\( (y + 4)^2 - 16 \) - Put it all together:
\( (x - 3)^2 - 9 + (y + 4)^2 - 16 = -9 \)
\( (x - 3)^2 + (y + 4)^2 = 16 \)
Result: The centre is \( (3, -4) \) and the radius is \( \sqrt{16} = 4 \).
Common Mistake: Forgetting to square the "halved" number when subtracting it. Always remember: \( (x + \text{half})^2 - (\text{half})^2 \).
3. Geometric Properties of Circles
There are three specific rules (or "theorems") from your GCSE days that are very useful in Coordinate Geometry. You need to know how to apply them to find gradients and equations.
Property 1: The Tangent and the Radius
The radius of a circle at any point is perpendicular (at a 90-degree angle) to the tangent at that same point.
- Why this matters: If you know the gradient of the radius (\( m_1 \)), the gradient of the tangent (\( m_2 \)) will be the negative reciprocal.
- Formula: \( m_1 \times m_2 = -1 \).
Property 2: Chords and Perpendicular Bisectors
If you draw a straight line from the centre of the circle that hits a chord (a line segment across the circle) at 90 degrees, it will bisect the chord (cut it exactly in half).
- Strategy Tip: The perpendicular bisector of any chord always passes through the centre of the circle. If you have two chords and find their perpendicular bisectors, where they cross is the centre!
Property 3: Angle in a Semicircle
Any angle drawn from the ends of a diameter to a point on the circumference is always 90 degrees.
- How to use it: If you have a triangle inside a circle and you can prove two sides are perpendicular (using gradients), then the third side (the hypotenuse) must be the diameter.
Did you know?
The property that the angle in a semicircle is a right angle is often called Thales's Theorem, named after an ancient Greek philosopher who lived over 2,500 years ago!
4. Intersections: Circles and Lines
In your exams, you might be asked to find where a line crosses a circle or if they even touch at all.
How to find intersection points:
- Take the equation of the straight line (e.g., \( y = mx + c \)).
- Substitute this into the circle equation \( (x - a)^2 + (y - b)^2 = r^2 \).
- Expand everything and simplify. You will end up with a quadratic equation in the form \( Ax^2 + Bx + C = 0 \).
- Solve the quadratic to find the \( x \)-coordinates, then plug them back into the line equation to find the \( y \)-coordinates.
Using the Discriminant (\( b^2 - 4ac \)):
You can tell how many times a line hits a circle without actually finding the points:
- If \( b^2 - 4ac > 0 \): The line crosses at two points (it's a chord).
- If \( b^2 - 4ac = 0 \): The line touches at one point (it's a tangent).
- If \( b^2 - 4ac < 0 \): The line does not meet the circle at all.
Key Takeaway: Treat intersections like a puzzle. Substituting the line into the circle is the "key" that unlocks the quadratic equation.
Summary: The "Cheat Sheet" for Circles
1. The Circle Formula: \( (x - a)^2 + (y - b)^2 = r^2 \).
2. Finding the Centre: Use Completing the Square.
3. Tangents: Gradient of tangent is the negative reciprocal of the radius gradient.
4. Chords: The perpendicular bisector of a chord goes through the centre.
5. Intersections: Substitute the line into the circle and use the discriminant \( b^2 - 4ac \).
Don't be afraid to sketch the circle! A quick drawing often helps you spot which geometric property you need to use. You've got this!