Welcome to the World of Circles!
In this chapter of PP1: Pure Maths, we are going to explore one of the most perfect shapes in nature: the circle. Whether you are looking at a pizza, a clock face, or a planet’s orbit, the math behind circles is everywhere! By the end of these notes, you’ll be able to find the center of any circle, calculate its size, and even find the equations of lines that just touch its edge.
Don't worry if this seems tricky at first! We will break it down into small, manageable steps. If you can use Pythagoras' Theorem and Completing the Square, you’re already halfway there!
1. The Standard Equation of a Circle
A circle is simply a set of points that are all the exact same distance (the radius) from a fixed middle point (the centre).
The Formula
The equation of a circle with centre \((a, b)\) and radius \(r\) is written as:
\[(x - a)^2 + (y - b)^2 = r^2\]
Think of it this way: This formula is actually just Pythagoras' Theorem in disguise! It calculates the distance between any point \((x, y)\) and the center \((a, b)\).
Important Tip: The "Sign Switch"
When you look at the equation, the signs inside the brackets are the opposite of the centre coordinates.
Example: If the equation is \((x - 3)^2 + (y + 5)^2 = 16\):
• The centre is \((3, -5)\) (Notice how \(-3\) became \(+3\) and \(+5\) became \(-5\)).
• The radius is \(\sqrt{16} = 4\).
Common Mistake: Many students forget to square root the number on the right. Remember, the formula uses \(r^2\), so if you see \(25\), the radius is \(5\), not \(25\)!
Quick Review:
• Centre \((a, b)\) uses \((x-a)\) and \((y-b)\).
• The radius is the square root of the constant on the right.
2. Finding the Centre and Radius (Completing the Square)
Sometimes, the exam won't give you the nice, tidy equation above. Instead, they might give it to you in a "scrambled" form, like this:
\(x^2 + 4x + y^2 - 6y - 12 = 0\)
To find the centre and radius, we use a technique called Completing the Square. Here is a step-by-step guide:
Step 1: Group the \(x\) terms and \(y\) terms together.
\((x^2 + 4x) + (y^2 - 6y) = 12\)
Step 2: Complete the square for both.
• For \(x^2 + 4x\): Half of \(4\) is \(2\). So we write \((x + 2)^2 - 2^2\).
• For \(y^2 - 6y\): Half of \(-6\) is \(-3\). So we write \((y - 3)^2 - (-3)^2\).
Step 3: Put it all together and simplify.
\((x + 2)^2 - 4 + (y - 3)^2 - 9 = 12\)
\((x + 2)^2 + (y - 3)^2 = 12 + 4 + 9\)
\((x + 2)^2 + (y - 3)^2 = 25\)
Final Result: The centre is \((-2, 3)\) and the radius is \(\sqrt{25} = 5\).
Key Takeaway: To unscramble a circle equation, half the coefficients of \(x\) and \(y\), put them in brackets, and subtract their squares!
3. Geometric Properties of Circles
You will need to use these three classic geometry rules to solve coordinate geometry problems. They are like "shortcuts" to finding missing information.
Property A: The Semicircle Rule
The angle in a semicircle is a right angle (\(90^\circ\)).
If you draw a triangle inside a circle where one side is the diameter, the angle touching the edge of the circle will always be \(90^\circ\). This is useful because it lets you use the Pythagorean theorem or gradients of perpendicular lines.
Property B: The Chord Bisector
The perpendicular from the centre to a chord bisects the chord.
Imagine a straight line (a chord) cutting through the circle. If you draw a line from the centre that hits the chord at exactly \(90^\circ\), it will cut that chord perfectly in half.
Property C: The Tangent Rule
The tangent to a circle is perpendicular to the radius at its point of contact.
A tangent is a line that just skims the edge of the circle at one point. At that exact point, the radius and the tangent meet to form a perfect "L" shape (\(90^\circ\)).
Did you know? This property is why wheels work! The point where the tire touches the road is always perpendicular to the axle (the centre).
4. Tangents and Normals
In your exam, you might be asked to find the equation of a tangent (the line skimming the circle) or a normal (the line going straight through the centre and the point of contact).
How to find the Equation of a Tangent:
Step 1: Find the gradient of the radius.
Use the centre point and the point of contact. Use the formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
Step 2: Find the gradient of the tangent.
Since the tangent is perpendicular to the radius, its gradient is the negative reciprocal.
Memory Aid: Flip it and change the sign! (e.g., if the radius gradient is \(2\), the tangent gradient is \(-\frac{1}{2}\)).
Step 3: Use the line equation formula.
Plug your tangent gradient and the point of contact into: \(y - y_1 = m(x - x_1)\).
How to find the Equation of a Normal:
The normal is just the line that contains the radius! So, you simply use the gradient of the radius (from Step 1 above) and the point of contact in the line equation formula. No "flipping" required!
Quick Review Box:
• Tangent Gradient: Perpendicular to radius (\(-\frac{1}{m}\)).
• Normal Gradient: Same as the radius (\(m\)).
5. Translation of Circles
A translation just means sliding the circle to a new position without changing its size.
If you move a circle with centre \((a, b)\) by the vector \(\begin{bmatrix} h \\ k \end{bmatrix}\), the new centre becomes \((a+h, b+k)\).
The radius stays exactly the same.
Example: If you translate \((x-1)^2 + (y-2)^2 = 9\) by the vector \(\begin{bmatrix} 3 \\ -1 \end{bmatrix}\):
• New centre \(x = 1 + 3 = 4\)
• New centre \(y = 2 - 1 = 1\)
• New equation: \((x-4)^2 + (y-1)^2 = 9\)
Key Takeaway: Translations only change the \(a\) and \(b\) values in the formula. The \(r^2\) value never changes when you slide a circle!