Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Find the equation of a circle (given center/radius or 3 points).
• Find the center and radius from an equation.
• Solve problems involving tangents to circles.
• Find intersection points of a circle and a line.

Lesson summary

This week, we delve into the fascinating world of circles within the realm of analytical geometry. Circles are fundamental geometric shapes, and understanding their properties and equations is crucial not only for mathematical proficiency but also for various real-world applications. From understanding the coverage area of cellular towers to designing circular structures in architecture, the principles of circles are ever-present. Mastering circles in analytical geometry provides a solid foundation for more advanced topics like conic sections and calculus.

Lesson notes

2.1 The Standard Form of the Equation of a Circle The standard form of the equation of a circle with center (a, b) and radius r is: (x - a)² + (y - b)² = r² Why this form? This equation is derived from the distance formula. It states that for any point (x, y) on the circle, the distance between (x, y) and the center (a, b) is always equal to the radius r.

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

5. Solution: Using the standard form, we have: (x - 2)² + (y - (-3))² = 5² (x - 2)² + (y + 3)² = 25 2.2 The General Form of the Equation of a Circle The general form of the equation of a circle is: x² + y² + Cx + Dy + E = 0 where C, D, and E are constants.

Relationship to Standard Form: The general form is obtained by expanding the standard form and rearranging the terms. 2.3 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 for Completing the Square: Group the x terms and y terms together: (x² + Cx) + (y² + Dy) = -E Complete the square for the x terms: Add (C/2)² to both sides.

Complete the square for the y terms: Add (D/2)² to both sides. Rewrite the equation in the form (x - a)² + (y - b)² = r² Identify the center (a, b) and radius r.

Example 2: Find the center and radius of the circle given by the equation x² + y² - 4x + 6y - 12 =

0. Solution: Group the x and y terms: (x² - 4x) + (y² + 6y) = 12 Complete the square for x: (x² - 4x + 4) + (y² + 6y) = 12 + 4 Complete the square for y: (x² - 4x + 4) + (y² + 6y + 9) = 12 + 4 + 9 Rewrite in standard form: (x - 2)² + (y + 3)² = 25 Therefore, the center is (2, -3) and the radius is √25 = 5. 2.4 Finding the Equation of a Circle Given Three Points If three points on a circle are given, we can find the equation of the circle by substituting the coordinates of the points into the general form of the equation (x² + y² + Cx + Dy + E = 0) and solving the resulting system of three linear equations for C, D, and

E. Then, complete the square to convert to standard form.

Example 3: Find the equation of the circle passing through the points (0, 0), (2, 0), and (0, 2).

Solution: Substitute (0, 0) into x² + y² + Cx + Dy + E = 0: 0 + 0 + 0 + 0 + E = 0 => E = 0 Substitute (2, 0) into x² + y² + Cx + Dy = 0: 4 + 0 + 2C + 0 = 0 => C = -2 Substitute (0, 2) into x² + y² + Cx + Dy = 0: 0 + 4 + 0 + 2D = 0 => D = -2 Therefore, the equation is x² + y² - 2x - 2y =

0. Completing the square: (x² - 2x + 1) + (y² - 2y + 1) = 1 + 1 (x - 1)² + (y - 1)² = 2 The center is (1, 1) and the radius is √2. 2.5 Tangents to Circles A tangent to a circle is a line that touches the circle at exactly one point. The tangent is perpendicular to the radius at the point of tangency.

Finding the Equation of a Tangent: Find the gradient of the radius from the center of the circle to the point of tangency. The gradient of the tangent is the negative reciprocal of the gradient of the radius (since they are perpendicular). Use the point-gradient form of a line equation (y - y₁) = m(x - x₁) to find the equation of the tangent, where (x₁, y₁) is the point of tangency and m is the gradient of the tangent.

Example 4: Find the equation of the tangent to the circle (x - 1)² + (y - 2)² = 25 at the point (5, 5).

Solution: Center of the circle is (1, 2).

Gradient of the radius: mᵣ = (5 - 2) / (5 - 1) = 3/4 Gradient of the tangent: mₜ = -4/3 Equation of the tangent: (y - 5) = (-4/3)(x - 5) y - 5 = (-4/3)x + 20/3 y = (-4/3)x + 20/3 + 5 y = (-4/3)x + 35/3 2.6 Intersection of a Circle and a Line To find the points of intersection between a circle and a line, solve the equations simultaneously. This usually involves substituting the equation of the line into the equation of the circle and solving the resulting quadratic equation.

Example 5: Find the points of intersection between the circle x² + y² = 25 and the line y = x +

1. Solution: Substitute y = x + 1 into the equation of the circle: x² + (x + 1)² = 25 Expand and simplify: x² + x² + 2x + 1 = 25 => 2x² + 2x - 24 = 0 Divide by 2: x² + x - 12 = 0 Factorize: (x + 4)(x - 3) = 0 Solve for x: x = -4 or x = 3 Find the corresponding y values: If x = -4, y = -4 + 1 = -3 If x = 3, y = 3 + 1 = 4 Therefore, the points of intersection are (-4, -3) and (3, 4). Guided Practice (With Solutions)

Question 1: A circle has a center at (-1, 4) and passes through the point (2, 8). Determine the equation of the circle.