Welcome to the World of Circles!
In this chapter, we are moving beyond simple shapes and looking at Circles through the lens of coordinate geometry. Whether it’s the path of a satellite orbiting Earth or the design of a clock face, circles are everywhere. By the end of these notes, you’ll be able to describe any circle using a mathematical formula and find the equations of lines that just touch them.
Don't worry if this seems tricky at first! If you can use Pythagoras' Theorem and "complete the square," you already have the most important tools in your kit.
1. The Standard Equation of a Circle
Every circle has two defining features: its centre and its radius. If we know these, we can write its equation.
The standard form for a circle with centre \( (a, b) \) and radius \( r \) is:
\( (x - a)^2 + (y - b)^2 = r^2 \)
Breaking it down:
- \( (a, b) \): This is the coordinate of the centre. Notice the minus signs in the formula! If the centre is \( (3, 4) \), the equation uses \( (x - 3) \) and \( (y - 4) \).
- \( r \): This is the radius (the distance from the centre to the edge).
- \( r^2 \): A very common mistake is forgetting to square the radius. If the radius is 5, the equation ends in 25.
Analogy: The Anchor and the Rope
Imagine the centre \( (a, b) \) is an anchor dropped in a coordinate plane. The radius \( r \) is a piece of rope tied to it. As you walk around the anchor with the rope pulled tight, your feet trace out the circle. The equation simply describes every point your feet touch!
Quick Review:
What is the centre and radius of \( (x + 2)^2 + (y - 5)^2 = 16 \)?
Answer: The centre is \( (-2, 5) \) (flip the signs!) and the radius is \( \sqrt{16} = 4 \).
2. From "Messy" Equations to "Clean" Equations
Sometimes, the exam will give you an equation that looks like this:
\( x^2 + 4x + y^2 - 6y - 12 = 0 \)
This is still a circle, but it's "expanded." To find the centre and radius, we need to complete the square for both \( x \) and \( y \).
Step-by-Step Guide:
- Group the terms: Put the \( x \)'s together and the \( y \)'s together. Move the plain number to the other side.
\( (x^2 + 4x) + (y^2 - 6y) = 12 \) - Complete the square for \( x \): Take half of 4 (which is 2), square it (which is 4).
\( (x + 2)^2 - 4 \) - Complete the square for \( y \): Take half of -6 (which is -3), square it (which is 9).
\( (y - 3)^2 - 9 \) - Put it all together:
\( (x + 2)^2 - 4 + (y - 3)^2 - 9 = 12 \)
\( (x + 2)^2 + (y - 3)^2 = 12 + 4 + 9 \)
\( (x + 2)^2 + (y - 3)^2 = 25 \)
Key Takeaway: Now we can see the centre is \( (-2, 3) \) and the radius is \( \sqrt{25} = 5 \).
3. Essential Circle Properties
To solve harder coordinate geometry problems, you need to remember three "golden rules" from your earlier studies. These are vital for finding gradients and lengths.
- The Semicircle Rule: The angle at the circumference in a semicircle is always a right angle (\( 90^\circ \)). If you see a triangle where the longest side is the diameter, it’s a right-angled triangle!
- The Chord Rule: A perpendicular line drawn from the centre of a circle to a chord will always bisect (cut in half) that chord.
- The Tangent Rule: A tangent (a line that touches the circle at one point) is always perpendicular (\( 90^\circ \)) to the radius at that point.
Did you know?
The word "tangent" comes from the Latin word 'tangere', which means "to touch."
4. Tangents and Normals
You may be asked to find the equation of a tangent or a normal at a specific point on the circle.
How to find the Tangent Equation:
- Find the gradient of the radius. Use the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \) between the centre and the point on the circle.
- Find the gradient of the tangent. Since the tangent is perpendicular to the radius, use the "negative reciprocal" rule: \( m_{tangent} = -\frac{1}{m_{radius}} \).
- Use the straight-line formula: \( y - y_1 = m(x - x_1) \) with your tangent gradient and the point on the circle.
How to find the Normal Equation:
The normal is a line that is perpendicular to the tangent. Secret tip: The normal at any point on a circle always passes through the centre of the circle! So, just find the equation of the line passing through the point and the centre.
Common Mistake:
Students often forget to flip the gradient and change the sign when moving from the radius to the tangent. If the radius gradient is \( \frac{2}{3} \), the tangent gradient must be \( -\frac{3}{2} \).
5. Translations of Circles
A translation is just a slide. If you move a circle, its size (the radius) stays exactly the same, but its centre changes.
If the circle \( x^2 + y^2 = r^2 \) is translated by the vector \( \begin{pmatrix} f \\ g \end{pmatrix} \), the new centre becomes \( (f, g) \) and the new equation is:
\( (x - f)^2 + (y - g)^2 = r^2 \)
Key Takeaway Summary:
1. Use \( (x - a)^2 + (y - b)^2 = r^2 \) for the circle equation.
2. Complete the square to find the centre and radius from expanded forms.
3. Use perpendicular gradients (\( m_1 \times m_2 = -1 \)) to find tangents.
4. Remember that the radius and tangent meet at \( 90^\circ \).
You've got this! Practice identifying the centre and radius first, then move on to the "completing the square" problems. One step at a time!