Welcome to the World of Circles!
In this chapter, we are moving from the straight lines you’ve already mastered into the beautiful, symmetrical world of Circles. Circles are everywhere—from the wheels on a bike to the ripples in a pond. In coordinate geometry, a circle is simply a collection of points that are all the exact same distance (the radius) from a fixed middle point (the centre).
Don’t worry if you find some of the algebra a bit "loopy" at first; we’ll break it down step-by-step so you can handle any circle question the exam throws at you!
1. The Equation of a Circle
The most important thing to learn is how to write the "identity card" for a circle—its equation. Every circle is defined by two things: its Centre and its Radius.
The General Formula
The standard equation of a circle with centre \( (a, b) \) and radius \( r \) is:
\[ (x - a)^2 + (y - b)^2 = r^2 \]
Wait, where does this come from?
Think back to the Pythagoras’ Theorem or the Distance Formula. If you pick any point \( (x, y) \) on the edge of the circle, the distance from that point to the centre \( (a, b) \) must always be \( r \). We are just using Pythagoras to say: "the horizontal distance squared + the vertical distance squared = the radius squared."
Drawing and Writing Equations
- Example 1: A circle has centre \( (3, -2) \) and radius \( 5 \).
Substitute into the formula: \( (x - 3)^2 + (y - (-2))^2 = 5^2 \).
Simplified: \( (x - 3)^2 + (y + 2)^2 = 25 \). - Example 2: If you see the equation \( x^2 + y^2 = 49 \), the centre is at the origin \( (0, 0) \) and the radius is \( \sqrt{49} = 7 \).
Quick Review Tip: Always watch the signs! In the brackets, it is minus the coordinate. So, \( (x + 5) \) actually means the x-coordinate of the centre is \( -5 \). Think of it as the "Opposite Rule."
Key Takeaway: To find the equation of a circle, you only need the centre and the radius. If you have those, you have the circle!
2. Expanding and Completing the Square
Sometimes, the exam won't give you the nice, neat bracket version. Instead, they might give you an expanded mess like this: \( x^2 + y^2 - 6x + 8y + 9 = 0 \).
To find the centre and radius from this, we use a technique called Completing the Square. Think of this as "re-packing" the equation back into its original brackets.
Step-by-Step Explanation:
- Group the terms: Put the \( x \)'s together and the \( y \)'s together. Move any plain numbers to the other side.
Example: \( (x^2 - 6x) + (y^2 + 8y) = -9 \) - Complete the square for \( x \): Halve the coefficient of \( x \), put it in a bracket squared, and subtract that number squared.
\( (x - 3)^2 - 9 \) - Complete the square for \( y \): Do the same for \( y \).
\( (y + 4)^2 - 16 \) - Clean up: Put it all together and move the "extra" numbers to the right.
\( (x - 3)^2 - 9 + (y + 4)^2 - 16 = -9 \)
\( (x - 3)^2 + (y + 4)^2 = 16 \)
Now we can see clearly: the Centre is \( (3, -4) \) and the Radius is \( \sqrt{16} = 4 \).
Key Takeaway: If the equation looks "unfolded," use Completing the Square to fold it back up and reveal the centre and radius.
3. Geometric Properties of Circles
Geometry isn't just about formulas; it's about the "rules" that circles must follow. These three properties are your best friends when solving tricky coordinate geometry problems:
Property 1: The Angle in a Semicircle
Rule: The angle subtended by a diameter at the circumference is always a right angle (\( 90^\circ \)).
How to use it: If you know two points on a circle form a \( 90^\circ \) angle with a third point, the line connecting the first two points must be the diameter. You can find the centre of the circle by finding the midpoint of that diameter!
Property 2: Perpendicular from Centre to Chord
Rule: A line drawn from the centre of the circle that is perpendicular to a chord will always bisect (cut in half) that chord.
How to use it: If you have the coordinates of a chord, find its midpoint. The line from the centre to that midpoint will have a gradient that is the negative reciprocal (\( m_1 m_2 = -1 \)) of the chord's gradient.
Property 3: Tangents and Radii
Rule: The radius of a circle is always perpendicular (\( 90^\circ \)) to the tangent at the point where they touch.
How to use it: This is a very common exam question!
- Find the gradient of the radius (from centre to the point on the edge).
- Find the "perpendicular gradient" (flip it and change the sign).
- Use \( y - y_1 = m(x - x_1) \) to find the equation of the tangent line.
Did you know? The word "Tangent" comes from the Latin word tangere, which means "to touch." A tangent just barely kisses the edge of the circle!
Key Takeaway: Whenever you see "tangent" or "chord," think "Perpendicular Gradients"! Use \( m_1 \times m_2 = -1 \).
4. Intersections: Do they meet?
In the coordinate plane, we often want to know if a line crosses a circle or if two circles bump into each other.
Line and Circle Intersection
To find where a line \( y = mx + c \) meets a circle, substitute the line's equation into the circle's equation. This will give you a quadratic equation.
- If the Discriminant \( (b^2 - 4ac) > 0 \): The line crosses at two points.
- If \( b^2 - 4ac = 0 \): The line touches at one point (it is a tangent!).
- If \( b^2 - 4ac < 0 \): The line misses the circle entirely.
Circle and Circle Intersection
To see if two circles intersect, compare the distance between their centres (\( d \)) to the sum of their radii (\( r_1 + r_2 \)).
- If \( d < r_1 + r_2 \): They overlap (intersect at two points).
- If \( d = r_1 + r_2 \): They touch exactly once on the outside.
- If \( d > r_1 + r_2 \): They are separate.
Common Mistake: When substituting a line into a circle, remember to expand the brackets carefully! \( (mx+c)^2 \) is not just \( m^2x^2 + c^2 \). Don't forget the middle term!
Key Takeaway: Use substitution to create a quadratic, then use the discriminant to see how many times they touch.
Final Quick Review
1. Equation: \( (x-a)^2 + (y-b)^2 = r^2 \) where \( (a,b) \) is the centre.
2. Finding Centre/Radius: Use Completing the Square if the equation is expanded.
3. Tangents: Gradient of radius \( \times \) Gradient of tangent \( = -1 \).
4. Diameter: The midpoint of the diameter is the centre of the circle.
5. Intersections: Substitute and check the discriminant \( b^2 - 4ac \).
You've got this! Circles are just a matter of finding the centre, knowing the radius, and using your straight-line skills for the rest. Keep practicing!