Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Determine the equation of a circle given its center and radius.
• Find the center and radius from a circle's equation.
• Find the equation of a tangent to a circle.
• Solve problems of intersection between a circle and a line.

Lesson summary

Analytical geometry provides a powerful way to describe and analyze geometric shapes using the coordinate plane. Specifically, understanding circles analytically is crucial for various applications, from engineering design to GPS navigation. This week, we'll focus on circles, exploring their standard equations, properties, and how to solve problems related to them. Analytical geometry, and circles in particular, can seem abstract, but it has direct relevance in sectors such as infrastructure development (e.g., designing circular intersections or drainage systems) and the gaming industry (e.g., programming the movement of characters or objects).

Lesson notes

2. 1.

The Standard Equation of a Circle: A circle is defined as the set of all points equidistant from a fixed point called the center. This distance is the radius. Let the center of the circle be the point (a, b) and the radius be 'r'. Consider any point (x, y) on the circle. By the distance formula, the distance between (x, y) and (a, b) is: √((x - a)² + (y - b)²) Since this distance is equal to the radius 'r', we have: √((x - a)² + (y - b)²) = r Squaring both sides gives us the standard equation of a circle: (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.

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. To find the center and radius from this general form, we need to complete the square for both the x and y terms. This process transforms the general form into the standard form.

Completing the Square: Consider the expression x² + Dx. To complete the square, we add and subtract (D/2)². Similarly, for y² + Ey, we add and subtract (E/2)². Let's see how this works.

Example 1: Finding the Center and Radius from the General Form Consider the equation: x² + y² - 4x + 6y - 23 = 0 Group the x and y terms: (x² - 4x) + (y² + 6y) = 23 Complete the square for the x terms: (x² - 4x + 4) + (y² + 6y) = 23 + 4 (We added (4/2)² = 4 to both sides)

Complete the square for the y terms: (x² - 4x + 4) + (y² + 6y + 9) = 23 + 4 + 9 (We added (6/2)² = 9 to both sides)

Rewrite as squared terms: (x - 2)² + (y + 3)² = 36 Identify the center and radius: Now the equation is in the standard form (x - a)² + (y - b)² = r².

Therefore, the center is (2, -3) and the radius is √36 = 6. 2.

3. Equation of a Tangent to a Circle: A tangent to a circle is a straight line that touches the circle at only one point. This point is called the point of tangency. The tangent is always perpendicular to the radius at the point of tangency.

Finding the Equation of a Tangent: Find the gradient of the radius (m_r): Given the center of the circle (a, b) and the point of tangency (x₁, y₁), the gradient of the radius is m_r = (y₁ - b) / (x₁ - a). Find the gradient of the tangent (m_t): Since the tangent is perpendicular to the radius, the product of their gradients is -

1. Therefore, m_t = -1 / m_r.

Use the point-slope form of a line: The equation of the tangent is given by y - y₁ = m_t(x - x₁), where (x₁, y₁) is the point of tangency and m_t is the gradient of the tangent.

Example 2: Finding the Equation of a Tangent Find the equation of the tangent to the circle x² + y² = 25 at the point (3, 4).

Identify the center and point of tangency: The center is (0, 0) and the point of tangency is (3, 4).

Find the gradient of the radius: m_r = (4 - 0) / (3 - 0) = 4/

3. Find the gradient of the tangent: m_t = -1 / (4/3) = -3/

4. Use the point-slope form: y - 4 = (-3/4)(x - 3)

Simplify: y - 4 = (-3/4)x + 9/4 => y = (-3/4)x + 25/4 Therefore, the equation of the tangent is y = (-3/4)x + 25/4 2.

4. Intersection of a Circle and a Straight Line: To find the point(s) of intersection between a circle and a straight line, we solve their equations simultaneously. This usually involves substitution.

Steps: Express one variable in terms of the other from the linear equation: Solve the equation of the line for either x or y. Substitute into the equation of the circle: Substitute the expression from step 1 into the equation of the circle. This will result in a quadratic equation in one variable.

Solve the quadratic equation: Solve the quadratic equation to find the value(s) of the variable. Substitute back to find the other variable: Substitute the value(s) found in step 3 back into the linear equation to find the corresponding value(s) of the other variable.

Interpret the solutions: The solutions represent the point(s) of intersection. If there are two distinct real solutions, the line intersects the circle at two points. If there is one real solution (repeated root), the line is tangent to the circle. If there are no real solutions, the line does not intersect the circle.

Example 3: Finding the Intersection Points Find the points of intersection of the circle x² + y² = 25 and the line y = x +

1. Express y in terms of x: y = x + 1 (already done) Substitute into the equation of the circle: x² + (x + 1)² = 25 Expand and simplify: x² + x² + 2x + 1 = 25 => 2x² + 2x - 24 = 0 => x² + x - 12 = 0 Solve the quadratic equation: (x + 4)(x - 3) = 0 => x = -4 or x = 3 Substitute back to find y: If x = -4, y = -4 + 1 = -3 => Point: (-4, -3) If x = 3, y = 3 + 1 = 4 => Point: (3, 4) Therefore, the points of intersection are (-4, -3) and (3, 4). Guided Practice (With Solutions)

Question 1: Determine the equation of a circle with center (-1, 2) and radius 4.