04Icircles

= Circles =

The rule ... ... ... ... ** x 2 + y 2 = r 2 **


 * creates a circle with
 * centre at (0, 0) ... ** (the origin) **
 * radius = r


 * A circle is a many-to-may relation
 * It fails the vertical line test
 * so a circle is __not__ a function


 * Example 1 **

Sketch .. x 2 + y 2 = 9


 * Solution:**
 * Can be written as: .. x 2 + y 2 = 3 2
 * So, this is a circle with
 * centre at (0, 0)
 * radius = 3


 * Domain is
 * x in [–3, 3]
 * Range is
 * y in [–3, 3]


 * Circles and Pythagoras' Theorem **
 * the similarity to Pythagoras' Theorem is not a coincidence
 * we can create a right-angled triangle inside the circle
 * with one vertex at the origin (0, 0) ,, and
 * another vertex on the circumference of the circle at (x, y)
 * the right angle will sit on the x-axis
 * the hypotenuse of the triangle will be the radius of the circle
 * the two shorter sides of the triangle will have lengths x and y
 * Apply Pythagoras' Theorem to the triangle
 * and we get x 2 + y 2 = r 2


 * Semi-Circle Functions **


 * If we make y the subject of the circle equation, we get:

math . \qquad \qquad x^2 + y^2 = r^2 \\.\\ . \qquad \qquad y^2 = r^2 - x^2 \\. \\ . \qquad \qquad y = \pm \sqrt{r^2 - x^2} math


 * so we get two equations which are functions

math . \qquad \qquad y = +\sqrt{r^2-x^2} \qquad \big\{ \text{the upper semicircle} \big\} \\.\\ . \qquad \qquad \text{and} \\.\\ . \qquad \qquad y = -\sqrt{r^2-x^2} \qquad \big\{ \text{the lower semicircle} \big\} math

math \text{Sketch} \quad y = \sqrt{16 - x^2} math
 * Example 2 **


 * Solution:**


 * This is the __upper__ half of a circle
 * centre at (0, 0)
 * radius = 4


 * Domain is
 * x in [0, 4]
 * Range is
 * y in [–4, 4]


 * Circles on the ClassPad Calculator **
 * The part of ClassPad we use for graphs requires you to enter a function
 * To draw circle, you will need to enter the two semicircles as two functions

math . \qquad \quad y1 = \sqrt{16-x^2} \quad and \quad y2 = -\sqrt{16-x^2} math


 * The circle will appear incomplete close to the x-axis
 * This is due to a limitation of the way the ClassPad draws functions


 * Other sections of the ClassPad can draw better circles
 * but those sections are beyond the scope of this course.


 * General Equation of a Circle **

The rule ... ... ... ... ** ( x – h) 2 + (y – k) 2 = r 2 **


 * creates a circle with
 * centre at (h, k)
 * radius = r


 * ie, the basic circle has been translated (shifted)
 * **h** units to the __right__ ... (opposite to the sign)
 * **k** units __up__ ... ... (opposite to the sign)


 * Example 3 **

Sketch (x –3) 2 + (y + 1) 2 = 4

and state the domain and range


 * Solution: **


 * Can be written as: .. (x –3) 2 + (y + 1) 2 = 2 2
 * So this is a circle with
 * centre at (3, –1)
 * radius = 2


 * Domain is
 * x in [1, 5]
 * Range is
 * y in [–3, 1]


 * Note **


 * This year, you will not be asked to find the x and y intercepts of a graph like the one in Example 3
 * But if you did have to find them, you would proceed in the normal way
 * To find the y-intercepts, let x = 0 and solve for y
 * To find the x-intercepts, let y = 0 and solve for x


 * Example 3 (continued) **

... ... ** y-intercepts **

math . \qquad \qquad \big( x - 3 \big)^2 + \big( y + 1 \big)^2 = 4 \\. \\ . \qquad \qquad \text{Let } x = 0 \\.\\ . \qquad \qquad \big( -3 \big)^2 + \big( y + 1 \big)^2 = 4 \\. \\ . \qquad \qquad 9 + \big( y + 1 \big)^2 = 4 \\. math math . \qquad \qquad \big( y + 1 \big)^2 = -5 \\. \\ . \qquad \qquad \text{no real solutions} \\. \\ . \\ . \qquad \qquad \text{so no y-intercepts} math

... ... ** x-intercepts **

math . \qquad \qquad \big( x - 3 \big)^2 + \big( y + 1 \big)^2 = 4 \\. \\ . \qquad \qquad \text{Let } y = 0 \\.\\ . \qquad \qquad \big( x - 3 \big)^2 + \big( 1 \big)^2 = 4 \\. \\ . \qquad \qquad \big( x - 3 \big)^2 + 1 = 4 \\. math math . \qquad \qquad \big( x - 3 \big)^2 = 3 \\. \\ . \qquad \qquad x - 3 = \pm \sqrt{3} \\. \\ . \qquad \qquad x - 3 = \pm \sqrt{3} \\. \\ . \qquad \qquad x = 3 \pm \sqrt{3} math .