Lesson Notes By Weeks and Term v5 - Grade 11
Analytical geometry – Week 10 focus
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Analytical geometry – Week 10 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 11
Term: 1st Term
Week: 10
Theme: General lesson support
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Analytical geometry is a powerful branch of mathematics that combines algebra and geometry. It provides a framework for describing geometric shapes and figures using algebraic equations and coordinate systems. This is crucial because it allows us to solve geometric problems using algebraic methods and vice versa. In the South African context, analytical geometry has applications in fields like surveying (land measurement), architecture (designing buildings), and town planning (laying out cities and infrastructure).
2.1 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 (x, y) represents any point on the circumference of the circle. (a, b) represents the coordinates of the center of the circle. r represents the radius of the circle (the distance from the center to any point on the circumference).
Derivation: This equation is derived from the distance formula. The distance between the center (a, b) and any point (x, y) on the circle must be equal to the radius r.
The distance formula is: √((x - a)² + (y - b)²) = r Squaring both sides gives us the standard equation: (x - a)² + (y - b)² = r² 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 General Form of the Equation of a Circle The equation of a circle can also be expressed in the general form: x² + y² + Dx + Ey + F = 0 Where D, E, and F are constants. To find the center and radius from the general form, we need to complete the square to rewrite the equation in standard form. This will be shown in an example. 2.2 Determining the Equation of a Circle Given the Center and Radius: Simply substitute the values of a, b, and r into the standard equation.
Given Three Points on the Circumference: This requires solving a system of equations. Substitute the coordinates of each point into the general form of the circle equation (x² + y² + Dx + Ey + F = 0). This will give you three equations with three unknowns (D, E, and F). Solve this system of equations to find the values of D, E, and F. Then, rewrite the equation in standard form by completing the square.
Given the Endpoints of a Diameter: Calculate the midpoint of the endpoints. This will be the center. Calculate the distance between the midpoint and one of the endpoints. This distance is the radius. 2.3 Intersection of a Circle and a Straight Line To find the points of intersection between a circle and a straight line, we need to solve their equations simultaneously. Solve the equation of the straight line for either x or y. Substitute this expression into the equation of the circle. This will give you a quadratic equation in one variable. Solve the quadratic equation. The solutions will be the x-coordinates (or y-coordinates) of the points of intersection. Substitute these values back into the equation of the straight line to find the corresponding y-coordinates (or x-coordinates).
Possible Outcomes: Two distinct solutions: The line intersects the circle at two points (secant).
One solution: The line is tangent to the circle (touches the circle at one point).
No solutions: The line does not intersect the circle. 2.4 Determining if a Point Lies Inside, Outside or On the Circle Given a circle equation (x - a)² + (y - b)² = r² and a point (p, q), substitute the point's coordinates into the left-hand side of the equation: If (p - a)² + (q - b)² r²: The point (p, q) lies outside the circle.
Example 1: Finding the equation of a circle given the center and radius.
A water tank is circular in shape. It is planned to position the center of the tank at point (2, -3) on a coordinate grid. The radius of the tank is to be 5 meters. Determine the equation that describes the boundary of the water tank.
Solution:
We are given the center (a, b) = (2, -3) and the radius r =
5.
Substitute these values into the standard equation:
(x - a)² + (y - b)² = r²
(x - 2)² + (y - (-3))² = 5²
(x - 2)² + (y + 3)² = 25
The equation of the circle is (x - 2)² + (y + 3)² = 25