Coordinate Geometry in the (x, y) Plane: Comprehensive Study Notes (Unit P2)
Hello mathematicians! Welcome to the essential chapter on Coordinate Geometry. While you covered the basics of straight lines in Unit P1, in P2, we take these skills and apply them to a fascinating new curve: the circle. This is where algebra meets geometry, and knowing these techniques is crucial for understanding later topics like differentiation (finding tangents!) and integration.
⭐ Quick Review: The Foundations (P1 Prerequisite)
Before diving into the circle, let’s quickly refresh the toolkit you need for the (x, y) plane:
Gradient (\(m\))
The gradient measures the steepness of a line.
Formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
Parallel Lines: Have the same gradient (\(m_1 = m_2\)).
Perpendicular Lines: The product of their gradients is \(-1\). \(m_1 \times m_2 = -1\).
Memory Aid: If a line has a gradient of \(2/3\), the perpendicular gradient is \(-3/2\). Flip it and negate it!
Distance and Midpoint
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Distance (\(d\)): Uses the Pythagorean theorem.
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
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Midpoint (\(M\)): Simply the average of the coordinates.
\(M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Key Takeaway from Review: Coordinate geometry relies on rearranging formulas and using basic algebra. Make sure your P1 basics are solid!
Section 1: The Equation of a Circle
The circle is the key geometric shape we study in depth in P2 coordinate geometry. All points on a circle are equidistant from a central point. That distance is the radius (\(r\)).
1.1 Standard Form of the Circle Equation
The standard equation is derived directly from the distance formula (Pythagoras Theorem).
The Formula:
If a circle has a centre \((a, b)\) and radius \(r\), the equation is:
\[(x - a)^2 + (y - b)^2 = r^2\]
- Finding the Centre: The coordinates \(a\) and \(b\) are the numbers being subtracted from \(x\) and \(y\). Note the sign change! If the equation is \((x + 3)^2\), then \(a = -3\).
- Finding the Radius: The value on the right-hand side (RHS) is \(r^2\). You must take the square root to find the radius \(r\).
Example: A circle has the equation \((x - 2)^2 + (y + 5)^2 = 49\).
Centre \((a, b)\) is \((2, -5)\).
Radius \(r\) is \(\sqrt{49} = 7\).
1.2 General Form and Completing the Square
Sometimes, the circle equation is expanded and looks messy. This is called the General Form:
\[x^2 + y^2 + 2gx + 2fy + c = 0\]
To find the centre and radius from the General Form, you must use the technique of Completing the Square, grouping the \(x\) terms and the \(y\) terms separately.
Step-by-Step: Completing the Square for Circles
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Group terms: Group the \(x\) terms and the \(y\) terms, and move the constant \(c\) to the RHS.
Example: \(x^2 + 4x + y^2 - 6y - 12 = 0\) becomes: \((x^2 + 4x) + (y^2 - 6y) = 12\) -
Complete the Square (x-terms): Take half of the coefficient of \(x\) and square it. Add this value to both sides of the equation.
Half of \(4\) is \(2\). \(2^2 = 4\).
\((x^2 + 4x + 4) + (y^2 - 6y) = 12 + 4\) -
Complete the Square (y-terms): Take half of the coefficient of \(y\) and square it. Add this value to both sides.
Half of \(-6\) is \(-3\). \((-3)^2 = 9\).
\((x^2 + 4x + 4) + (y^2 - 6y + 9) = 12 + 4 + 9\) -
Rewrite in Standard Form: Rewrite the bracketed quadratic expressions as squared terms.
\[(x + 2)^2 + (y - 3)^2 = 25\] -
Identify Centre and Radius:
Centre: \((-2, 3)\)
Radius: \(\sqrt{25} = 5\)
🚨 Common Mistake Alert: Students often forget to add the squared terms to the RHS of the equation. Remember, whatever you add to the left, you MUST add to the right to keep the equation balanced!
Key Takeaway: The Standard Form \((x - a)^2 + (y - b)^2 = r^2\) is your target. If you see the General Form, immediately complete the square!
Section 2: Circle Properties and Perpendicularity
Coordinate geometry problems involving circles often require you to use the relationship between the centre, the radius, and other lines.
2.1 Tangents and Radii
A tangent is a straight line that touches the circle at exactly one point. This point is called the point of contact.
The single most important property relating lines and circles is:
The radius drawn to the point of contact is always perpendicular to the tangent line at that point.
How to Find the Equation of a Tangent
Suppose you are given a circle and a point \(P\) on the circumference, and you need the equation of the tangent at \(P\).
- Find the Centre (C): Determine the coordinates \((a, b)\) from the circle’s equation.
- Find the Gradient of the Radius (CP): Use the gradient formula with the centre \(C\) and the point \(P\). Let this be \(m_{radius}\).
- Find the Gradient of the Tangent: Use the perpendicular rule: \(m_{tangent} = -1 / m_{radius}\).
- Find the Equation: Use the tangent gradient and the point \(P\) in the formula \(y - y_1 = m(x - x_1)\).
Analogy: Imagine spinning a ball on a string. If you let go, the ball flies off in a direction perpendicular to the string (the radius) at that instant. That flight path is the tangent!
2.2 Chords and Perpendicular Bisectors
A chord is a straight line segment whose endpoints both lie on the circle.
A second key property of circles is:
The perpendicular bisector of any chord passes through the centre of the circle.
If you have two points (A and B) on a circle, and you find the equation of the line that cuts the chord AB exactly in half (\(M\)) and is perpendicular to it, that line must pass through the centre \((a, b)\).
Why is this useful?
If you are given two endpoints of a diameter, the centre is just the midpoint. But if you are given three points on the circumference and asked to find the centre, you must:
- Find the perpendicular bisector of the first chord (e.g., AB).
- Find the perpendicular bisector of the second chord (e.g., BC).
- Solve the two resulting linear equations simultaneously. The intersection point is the centre of the circle.
Key Takeaway: Perpendicularity (\(m_1 m_2 = -1\)) is the fundamental link in circle geometry, governing the relationship between tangents, radii, chords, and the centre.
Section 3: Intersections of Lines and Circles (Simultaneous Equations)
A classic coordinate geometry problem is determining how many times a straight line intersects a circle.
Because both the circle equation (quadratic) and the line equation (linear) exist in the same plane, we solve them simultaneously by substitution.
The Process
- Isolate \(x\) or \(y\): Rearrange the linear equation (the straight line) to make either \(x\) or \(y\) the subject (e.g., \(y = mx + c\)).
- Substitute: Substitute the linear expression into the circle equation (the quadratic). This will eliminate one variable and leave you with a single, potentially complicated, quadratic equation (in terms of \(x\) or \(y\)).
- Simplify: Expand and collect all terms onto one side to get the standard quadratic form: \(Ax^2 + Bx + C = 0\).
- Use the Discriminant: The nature of the intersection depends on the value of the discriminant, \(\Delta = B^2 - 4AC\).
Don't worry if the substitution seems messy—just be meticulous with your expansion and collection of like terms!
3.1 Using the Discriminant (\(\Delta\))
The discriminant tells you how many real solutions (intersection points) exist for the resulting quadratic equation:
| Discriminant Value | Number of Intersections | Geometric Interpretation |
|---|---|---|
| \(\mathbf{B^2 - 4AC > 0}\) | Two distinct real roots | The line is a secant; it crosses the circle at two points. |
| \(\mathbf{B^2 - 4AC = 0}\) | One repeated real root | The line is a tangent; it touches the circle at exactly one point. |
| \(\mathbf{B^2 - 4AC < 0}\) | No real roots | The line misses the circle entirely. |
Tip for Tangent Problems: If a question asks you to show that a line is tangent to a circle, you must solve simultaneously and prove that \(B^2 - 4AC = 0\).
Did You Know?
The study of conic sections (circles, ellipses, parabolas, hyperbolas) dates back to ancient Greek mathematicians like Apollonius, who developed the concepts centuries before the invention of the coordinate plane! Descartes and Fermat formalised coordinate geometry in the 17th century.
Summary and Quick Review
Mastering coordinate geometry in P2 means mastering the circle and its perpendicular relationships.
Quick Check: Key Formulas
- Standard Circle: \((x - a)^2 + (y - b)^2 = r^2\)
- Perpendicular Gradients: \(m_1 \times m_2 = -1\)
- Discriminant: \(\Delta = B^2 - 4AC\)
If you can confidently switch between the General Form and the Standard Form (using completing the square), and use the perpendicular rule to relate radii and tangents, you are set for success in this chapter! Good luck!