Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Determine the equation of a circle from its center and radius.
• Find the center and radius from a circle's equation.
• Sketch the graph of a circle.
• Solve problems involving tangents to circles.
• Find the points of intersection between a circle and a line.

Lesson summary

Analytical geometry is the study of geometry using a coordinate system and the principles of algebra. Specifically, we will focus on the geometry of circles. Understanding circles is not just an academic exercise. From the design of roundabouts improving traffic flow in urban areas like Johannesburg to understanding the propagation range of cell towers connecting rural communities, circles play a vital role in South African infrastructure and daily life.

Furthermore, understanding circles is crucial for fields like engineering (designing gears and wheels), architecture (domes and circular structures), and even telecommunications (signal coverage areas).

Lesson notes

2.1 The Equation of a Circle The standard equation of a circle with center (a; b) and radius r is: (x - a)² + (y - b)² = r² Center (a; b): This point is equidistant from all points on the circle.

Radius (r): This is the distance from the center of the circle to any point on the circle.

Why this formula works: The formula is derived directly from the distance formula (Pythagorean theorem). Consider any point (x; y) on the circle. The distance between (x; y) and the center (a; b) must be equal to the radius r.

Therefore: √[(x - a)² + (y - b)²] = r Squaring both sides gives us the standard equation of a circle.

Special Case: Circle Centered at the Origin If the center of the circle is at the origin (0; 0), the equation simplifies to: x² + y² = r² 2.2 General Form of the Equation of a Circle The general form of the equation of a circle is: x² + y² + Dx + Ey + F = 0 Where D, E, and F are constants. Converting from General Form to Standard Form To find the center and radius from the general form, you must complete the square for both x and y.

Here's how: Group the x terms and y terms together: (x² + Dx) + (y² + Ey) = -F Complete the square for the x terms: Take half of the coefficient of x (which is D/2), square it (D²/4), and add it to both sides.

Complete the square for the y terms: Take half of the coefficient of y (which is E/2), square it (E²/4), and add it to both sides.

Rewrite the equation in standard form: (x + D/2)² + (y + E/2)² = -F + D²/4 + E²/4 Identify the center and radius: The center is (-D/2; -E/2) and the radius is √(-F + D²/4 + E²/4). Note that (-F + D²/4 + E²/4) must be greater than zero for the equation to represent a real circle. 2.3 Tangents to Circles A tangent to a circle is a line that touches the circle at exactly one point. This point is called the point of tangency.

Key Property: The tangent to a circle is perpendicular to the radius at the point of tangency. Finding the Equation of a Tangent Find the gradient of the radius: Given the center (a; b) and the point of tangency (x₁; y₁), the gradient of the radius is (y₁ - b) / (x₁ - a).

Find the gradient of the tangent: Since the tangent is perpendicular to the radius, the gradient of the tangent (m T ) is the negative reciprocal of the gradient of the radius (m R ): m T = -1/m R .

Use the point-slope form of a line: The equation of the tangent is y - y₁ = m T (x - x₁). 2.4 Intersection of a Circle and a Line To find the points of intersection between a circle and a line, solve the equations simultaneously. Express one variable in terms of the other: From the equation of the line (e.g., y = mx + c), express y in terms of x (or x in terms of y). Substitute into the equation of the circle: Substitute the expression from step 1 into the equation of the circle. This will give you a quadratic equation in one variable (either x or y).

Solve the quadratic equation: Solve the quadratic equation to find the values of the variable.

Find the corresponding values: Substitute the values obtained in step 3 back into either the equation of the line or the equation of the circle to find the corresponding values of the other variable. The solutions are the points of intersection: The solutions (x; y) are the coordinates of the points where the line intersects the circle. Important

Note: If the quadratic equation has two distinct real roots, the line intersects the circle at two points (secant). If the quadratic equation has one real root (repeated root), the line is tangent to the circle (intersects at one point). If the quadratic equation has no real roots, the line does not intersect the circle. 2.5 Worked

Examples: Example 1: Find the equation of a circle with center (2; -3) and radius

4. Solution: Using the standard equation (x - a)² + (y - b)² = r², we have: (x - 2)² + (y - (-3))² = 4² (x - 2)² + (y + 3)² = 16 Example 2: Find the center and radius of the circle given by the equation: x² + y² - 4x + 6y - 12 =

0. Solution: We complete the square: (x² - 4x) + (y² + 6y) = 12 (x² - 4x + 4) + (y² + 6y + 9) = 12 + 4 + 9 (x - 2)² + (y + 3)² = 25 = 5² Therefore, the center is (2; -3) and the radius is

5. Example 3: Find the equation of the tangent to the circle x² + y² = 25 at the point (3; 4).

Solution: The center of the circle is (0; 0).

Gradient of the radius: m R = (4 - 0) / (3 - 0) = 4/3 Gradient of the tangent: m T = -1 / (4/3) = -3/4 Equation of the tangent: y - 4 = (-3/4)(x - 3) y - 4 = (-3/4)x + 9/4 y = (-3/4)x + 9/4 + 16/4 y = (-3/4)x + 25/4 Example 4: Find the points of intersection of the circle x² + y² = 16 and the line y = x + 2.