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 centre and radius.
• Find the centre and radius from a circle's equation.
• Solve problems involving circles, tangents, and chords.
• Apply knowledge of circles to solve contextual problems.

Lesson summary

Analytical geometry provides a powerful way to describe geometric shapes using algebraic equations. Focusing on circles in Grade 12 builds on the foundation laid in earlier grades and provides essential tools for various applications. Circles are fundamental shapes appearing everywhere in our world, from wheels and gears to satellite orbits and cellular structures. Understanding their mathematical representation is crucial for engineering, physics, computer graphics, and even understanding the distribution of resources like cell phone towers across South Africa.

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 centre. This fixed distance is called the radius. Let the centre of the circle be (a; b) and the radius be r. Consider any point (x; y) on the circle. Using the distance formula, the distance between (x; y) and (a; b) must be equal to r.

Therefore: √[(x - a)² + (y - b)²] = r Squaring both sides, we get the standard equation of a circle: (x - a)² + (y - b)² = r² Where: (a; b) is the centre of the circle. r is the radius of the circle.

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

4. Solution: Using the standard equation: (x - a)² + (y - b)² = r² Substitute a = 2, b = -3, and r = 4: (x - 2)² + (y - (-3))² = 4² (x - 2)² + (y + 3)² = 16 2.2 The General Equation of a Circle The general equation of a circle is derived from the standard equation by expanding and rearranging the terms.

Starting with the standard form: (x - a)² + (y - b)² = r² Expanding: x² - 2ax + a² + y² - 2by + b² = r² Rearranging: x² + y² - 2ax - 2by + a² + b² - r² = 0 Let D = -2a, E = -2b, and F = a² + b² - r² Then the general equation becomes: x² + y² + Dx + Ey + F = 0 Important

Note: Not every equation in this form represents a circle. For it to be a circle, the radius squared, r², which is a² + b² - F, must be greater than zero. If a² + b² - F t be the tangent's gradient and m r be the radius's gradient. Then 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), where (x₁; y₁) is the point of tangency.

Example 3: Find the equation of the tangent to the circle (x - 1)² + (y - 2)² = 25 at the point (5; 5).

Solution: Centre of the circle: (1; 2)

Gradient of the radius: (5 - 2) / (5 - 1) = 3 / 4 Gradient of the tangent: -1 / (3/4) = -4/3 Equation of the tangent: y - 5 = (-4/3)(x - 5)

Simplify: y = (-4/3)x + 20/3 + 5 y = (-4/3)x + 35/3 Guided Practice (With Solutions)

Question 1: Determine the equation of a circle with centre (-1; 4) and radius √

5. Solution: Using the standard equation (x - a)² + (y - b)² = r²: Substitute a = -1, b = 4, and r = √5: (x - (-1))² + (y - 4)² = (√5)² (x + 1)² + (y - 4)² = 5 Question 2: The equation of a circle is given by x² + y² + 8x - 2y + 8 =

0. Determine the coordinates of the centre and the length of the radius.

Solution: Group terms: (x² + 8x) + (y² - 2y) = -8 Complete the square for x: (x² + 8x + 16) + (y² - 2y) = -8 + 16 Complete the square for y: (x² + 8x + 16) + (y² - 2y + 1) = -8 + 16 + 1 Rewrite: (x + 4)² + (y - 1)² = 9 Therefore: Centre: (-4; 1)

Radius: √9 = 3 Question 3: Find the equation of the tangent to the circle x² + y² = 25 at the point (3; -4).

Solution: Centre of the circle: (0; 0)

Gradient of the radius: (-4 - 0) / (3 - 0) = -4/3 Gradient of the tangent: -1 / (-4/3) = 3/4 Equation of the tangent: y - (-4) = (3/4)(x - 3)

Simplify: y + 4 = (3/4)x - 9/4 y = (3/4)x - 9/4 - 16/4 y = (3/4)x - 25/4 Independent Practice (Questions Only) Determine the equation of a circle with centre (0; -5) and radius

7. Find the centre and radius of the circle given by the equation x² + y² - 6x + 4y - 12 =

0. The equation of a circle is (x - 3)² + (y + 1)² =

1

6. Does the point (7; 0) lie inside, outside, or on the circle? Find the equation of the tangent to the circle (x + 2)² + (y - 3)² = 25 at the point (1; 7). Determine the equation of the circle that passes through the point (2; 4) and has its centre at (0; 0). A circle has the equation x² + y² + 2x - 8y + k =

0. Determine the value of k if the radius of the circle is

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

1. Find the equation of the circle that has the points (1; 2) and (5; 2) as the ends of a diameter. Find the length of the tangent from the point (5,4) to the circle x² + y² + 2x - 4y - 20 =

0. Show that the equation x² + y² - 4x + 6y + 15 = 0 represents no real points.