Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Find the equation of a circle given its center and radius.
• Find the center and radius from a circle's equation.
• Solve problems involving tangents to circles.
• Find the points of intersection between a circle and a line.
• Apply these principles to solve problems.

Lesson summary

Analytical geometry is a powerful tool that allows us to describe and analyze geometric shapes using algebraic equations. This week, we focus specifically on circles. Understanding circles is crucial not just for further mathematics, but also for applications in surveying, engineering (think circular roads or structures), design, and even fields like astronomy where understanding orbits is vital. Many everyday objects, from wheels to stadium designs, incorporate circular geometry. In a South African context, consider the design of roundabouts for traffic management or the layout of circular cultural villages.

Lesson notes

2.1 The Equation of a Circle: The standard equation of a circle with center (a, b) and radius r is given by: (x - a)² + (y - b)² = r² (a, b) represents the coordinates of the center of the circle. r represents the radius of the circle. (x, y) represents any point on the circumference of the circle.

Why this equation works: This equation is derived directly from the Pythagorean theorem. Imagine a right-angled triangle formed by the radius, a horizontal line from the center to a point (x, y) on the circle, and a vertical line from (x, y) back to the horizontal line. The length of the horizontal line is |x - a|, the length of the vertical line is |y - b|, and the length of the hypotenuse is r.

Therefore, by the Pythagorean theorem, (x - a)² + (y - b)² = r².

Special Case: Circle centered at the origin (0, 0): When the center of the circle is at the origin, the equation simplifies to: x² + y² = r² 2.2 The General Form of the Equation of a Circle: The general form of the equation of a circle is given by: x² + y² + Dx + Ey + F = 0 Where D, E, and F are constants. Converting from General Form to Standard Form (Completing the Square): To find the center and radius of a circle given in general form, we need to convert it to standard form by completing the square.

Steps: Group the x terms and y terms: (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 of the equation. (x² + Dx + D²/4) + (y² + Ey) = -F + D²/4 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 of the equation. (x² + Dx + D²/4) + (y² + Ey + E²/4) = -F + D²/4 + E²/4 Rewrite the expressions in parentheses as squared terms: (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). Important

Note: If -F + D²/4 + E²/4 is negative, then the equation does not represent a circle because the radius cannot be the square root of a negative number. 2.3 Tangents to Circles: A tangent to a circle is a line that touches the circle at only 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: Calculate the gradient (m r ) of the line segment connecting the center of the circle and the point of tangency.

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 t = -1/m r .

Use the point-slope form of a line: Use the point of tangency (x 1 , y 1 ) and the gradient of the tangent (m t ) to find the equation of the tangent using the formula: y - y 1 = m t (x - x 1 ). 2.4 Intersection of a Circle and a Line: To find the points where a line intersects a circle, we need to solve the equations of the line and the circle simultaneously.

Steps: Solve the equation of the line for either x or y. Substitute the expression for x or y into the equation of the circle. Solve the resulting quadratic equation for the remaining variable. This will give you the x or y coordinates of the points of intersection. Substitute the values you found in step 3 back into the equation of the line to find the corresponding coordinates.

Possible Outcomes: Two distinct solutions: The line intersects the circle at two points (the line is a secant).

One solution: The line is tangent to the circle (it touches the circle at one point).

No solutions: The line does not intersect the circle.

Worked example

Example 1: Find the equation of the 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