Section 1.4 Circles. Objective #1: Writing the Equation of a Circle in Standard Form.

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1 1 Section 1. Circles Objective #1: Writing the Equation of a Circle in Standard Form. We begin by giving a definition of a circle: Definition: A Circle is the set of all points that are equidistant from a fixed point (h, k) called the center in the xy-plane. The radius of the circle, r, is the length between the center to a point on the circle. (x, y) To derive the equation of a circle, let (x, y) represent any point on the circle. To find the r distance between this point and the center in the xy-plane, we can use the distance formula: (h, k) d = (x x 1 ) +(y y 1 ) But, since (x 1, y 1 ) = (h, k), (x, y ) = (x, y) and d = r, we get: r = (x h) +(y k) Squaring both sides and rearranging yields the standard form of the equation of a circle: (x h) + (y k) = r If we were to expand, combine like terms, and get zero on one side of the equation, this would produce the general form of the equation of a circle. Definition: Given a circle of radius r centered at (h, k), then 1) The standard form of the equation of the circle is: (x h) + (y k) = r Note that if the origin (0, 0) is the center, then the equation becomes x + y = r ) The general form of the equation of the circle is: x + y + ax + by + c = 0 where a, b, and c are constants. 3) The unit circle is a circle with radius 1 centered at the origin: x + y = 1 Given the radius and the center, write a) the standard form and b) general form of each equation: Ex. 1 r = 7, (h, k) = ( 3, )

2 a) Replace r with 7 and (h, k) with ( 3, ) in the standard form: (x h) + (y k) = r (x ( 3)) + (y ()) = 7 (simplify) (x + 3) + (y ) = 9 Standard form b) To find the general form, expand, combine like terms, and get zero on one side: (x + 3) + (y ) = 9 (x + 3)(x + 3) + (y )(y ) = 9 (FOIL) x + 6x y y + = 9 (combine like terms) x + y + 6x y + 13 = 9 (subtract 9 from both sides) x + y + 6x y 36 = 0 General form Ex. r = 5, (h, k) = ( 1, 7 8 ) a) Replace r with 5 and (h, k) with ( 1, 7 ) in the standard form: 8 (x h) + (y k) = r (x ( 1 )) + (y ( )) = ( ) (simplify) (x 1 ) + (y ) = 5 Standard form b) To find the general form, expand, combine like terms, and get zero on one side: (x 1 ) + (y ) = 5 (x 1 )(x 1 ) + (y )(y ) = 5 (FOIL) x x y + 7 y = 5 x + y x + 7 y = 5 x + y x + 7 y 35 6 Objective #: Graphing a Circle (combine like terms) (subtract 5 = 0 General form from both sides) If the equation of the circle is in standard form, we simply determine the center (h, k) and the radius r. We then locate the center of the circle on the graph paper and plot four points r units above, below, to the left, and to the right of the center. Finally, we draw a circle through those four points.

3 3 In the case where the equation is not in standard form, we will first need to rewrite the equation in standard form. This may include the need to complete the square in order to get the equation in standard form. Find a) the center and the radius of the circle, b) sketch the graph of the circle, and c) find the intercepts: Ex. 3 6(x ) + 6(y + 1) = 5 a) To get the equation in standard form, divide both sides by 6. This will yield: (x ) + (y + 1) = 9 Thus, (h, k) = (, 1) and r = 3. b) To graph the circle, we locate the center (, 1) and then plot four points that are 3 units above, below, to the left, and to the right of (, 1). Finally, we draw a circle through those four points (see figure on the right). c) To find the x-intercepts, let y = 0 and solve: (x ) + (0 + 1) = 9 (x ) + 1 = 9 (subtract 1 from both sides) (x ) = 8 (use the square root property) x = ± 8 (add to both sides and simplify) x = ± 0.88 or.88 To find the y-intercepts, let x = 0 and solve: (0 ) + (y + 1) = 9 + (y + 1) = 9 (subtract from both sides) (y + 1) = 5 (use the square root property) y + 1 = ± 5 (subtract 1 from both sides) y = 1 ± or 1.36 x-intercepts: (, 0) ( 0.88, 0) & ( +, 0) (.88, 0) y-intercepts: (0, 1 5 ) (0, 3.36) & (0, ) (0, 1.36)

4 Ex. 7x + 7y 8x + 8y + 8 = 0 a) First, divide both sides by seven: x + y 1x + y + = 0 (rearrange terms) x 1x + y + y + = 0 x 1x + y + y = Now, complete the square: x 1 1x + ( ) + y + y + ( (subtract from both sides) ) = + ( 1 ) + ( ) x 1x y + y + = (factor & simplify) (x 6) + (y + ) = Thus, (h, k) = (6, ) and r = b) To graph the circle, we locate the center (6, ) and then plot four points that are units above, below, to the left, and to the right of (6, ). Finally, we draw a circle through those four points (see figure on the right). c) To find the x-intercepts, let y = 0 and solve: (x 6) + (0 + ) = (x 6) + = (subtract from both sides) (x 6) = 1 (use the square root property) x 6 = ± 1 (add 6 to both sides and simplify) x = 6 ± or 9.6 To find the y-intercepts, let x = 0 and solve: (0 6) + (y + ) = 36 + (y + ) = (subtract 36 from both sides) (y + ) = 0 (use the square root property) y + = ± 0 which has no real solution x-intercepts: (6 3, 0) (.536, 0) & (6 + 3, 0) (9.6, 0) y-intercepts: None

5 5 Find the standard form of the equation of each circle: Ex. 5 A circle with endpoints of a diameter at ( 3, 1) and (7, 5). The center of the circle is the midpoint between the endpoints of the diameter. Midpoint = ( x 1 + x, y 1+ y ) = ( 3+7, 1+5 ) = (, 3) Thus, (h, k) = (, 3). The radius of the circle is the distance from the center to one of the points on the circle: r = (x x 1 ) +(y y 1 ) (use (, 3) and (7, 5)) = (7 ) +(5 3) (simplify) = 5+ = 9 Hence, r = 9. Plugging in (h, k) and r, the equation becomes: (x ) + (y 3) = 9 Ex. 6 The graph of the circle is given: The circle is centered at (, 1), so (h, k) = (, 1) and the distance from the center to (, 1) on the circle is, so r =. Plugging in (h, k) and r, the equation becomes: (x + ) + (y 1) = Note, to graph a circle on a graphing calculator, you will need to first solve for y. You will get two expressions, a positive square root and a negative square root. You will then need to let Y1 = 1 st expression and Y = nd expression and graph both at the same time to get a graph of a circle.

King Fahd University of Petroleum and Minerals Prep-Year Math Program Math (001) - Term 181 Recitation (1.1)

King Fahd University of Petroleum and Minerals Prep-Year Math Program Math (001) - Term 181 Recitation (1.1) Recitation (1.1) Question 1: Find a point on the y-axis that is equidistant from the points (5, 5) and (1, 1) Question 2: Find the distance between the points P(2 x, 7 x) and Q( 2 x, 4 x) where x 0. Question