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Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Do Now: What is the locus of points 4 units from a given point? The locus is a circle whose center.

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Presentation on theme: "Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Do Now: What is the locus of points 4 units from a given point? The locus is a circle whose center."— Presentation transcript:

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2 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Do Now: What is the locus of points 4 units from a given point? The locus is a circle whose center is the given point and that has a radius of 4 units Aim: How do we find the equation of the locus of points at a given distant from a given point?

3 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Describe the locus of points 3 units from the from the origin. The locus of points that are at distance d from fixed point A is a circle whose center is point A and the length of whose radius is distance d. The locus is a circle with a radius of 3 units and whose center is the origin (0,0) 3 What is the equation of this locus? Circle Basics

4 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Point P(x,y) is on circle O whose radius is five units. Use the distance formula to find the equation for the circle. (0,0) 5 P(x,y) (4,-3)(3,-4)(0,-5) (-3,-4) (-4,-3) (-5,0)(5,0) (-4,3) (-3,4) (0,-5) P(3,4) (4,3) When r = the radius of a circle, then the equation of a circle whose center is the origin (0,0) and whose radius has a length of r is the equation x 2 + y 2 = r = 5 2 = 25

5 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle The equation x 2 + y 2 = 100 represents the locus of all points at a given distance from the origin. What is the distance? When r = the radius of a circle, then the equation of a circle whose center is the origin (0,0) and whose radius has a length of r is the equation x 2 + y 2 = r 2 r 2 = 100 r = 10 radius measures 10 units Write an equation of the locus of points that are at a distance of 5 units from the origin. x 2 + y 2 = r 2 x 2 + y 2 = 25 r = 5 Model Problem

6 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Describe fully the locus of points for the given equation. A) x 2 + y 2 = 45 B) 4x 2 + 4y 2 = 64 The locus is a circle whose center is origin and whose radius is the square root of 45 or The locus is a circle whose center is origin and whose radius of 4. 4 x 2 + y 2 = 16 Is (8,6) located on the locus of a points whose distance from the origin is 10? x 2 + y 2 = = = 100 YES

7 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Write an equation for the locus of points that are equidistant from the circles whose equations are x 2 + y 2 = 16 and x 2 + y 2 = 64. x 2 + y 2 = 16 is a circle with center at the origin and with a radius of 4 units. x 2 + y 2 = 64 is a circle with center at the origin and with a radius of 8 units. The locus of points would therefore be a circle whose radius is halfway between 4 and 8 units from the origin, or 6 units from the origin and whose equation is x 2 + y 2 = 36. Model Problem

8 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Point (x, y) is on a circle whose radius is r units and whose center is the point (h, k). Use the distance formula to find the equation for the circle. r (h,k)(h,k) (x,y)(x,y) (x – h) 2 + (y – k) 2 = r 2 What is the value of (h, k)? (4,3)(4,3) (4, 3) What is the equation for this circle whose radius is 5 with center at (4, 3)? (x – 4) 2 + (y – 3) 2 = 5 2 = 5 (x – 4) 2 + (y – 3) 2 = 25 The standard form of an equation of a circle with center (h, k) and radius r is ( x – h) 2 + (y – k) 2 = r 2

9 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Circle with Center (h,k) The standard form of an equation of a circle with center (h, k) and radius r is (x – h) 2 + (y – k) 2 = r 2 What is the equation for the circle whose radius is 5 with center at (4, 3)? (x – 4) 2 + (y – 3) 2 = r= 5 (4,3) (h,k)(h,k) O x 2 + y 2 = 25 (0,0) O (x – 4) 2 + (y – 3) 2 = 5 2 is a translation of x 2 + y 2 = 25 under the rule T 4,3 (x,y) = (x + 4, y + 3), or

10 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Write and equation of the locus of points 6 units from the point (-3,5). Standard equation of circle (x – h) 2 + (y – k) 2 = r 2 h = -3, k = 5, r = 6 (x – -3) 2 + (y – 5) 2 = 6 2 (x + 3) 2 + (y – 5) 2 = 36 Find the center and radius of the circle whose equation is (x + 5) 2 + y 2 = 25. Does the origin lie on the locus of the points represented by the equation? Center - (-5,0) Radius = 5 (0 + 5) = 25 Origin -(0,0) (5) 2 = 25 YES Model Problem

11 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Find the coordinates of the center of the circle whose equation is x x + y 2 + 2y = -40 Standard equation of circle (x – h) 2 + (y – k) 2 = r 2 x x + ( )+ y 2 + 2y + ( )= -40 complete the squares (x + 7) 2 + (y + 1) 2 = 10 rewrite as squares of binomials center: (h, k) = (-7, -1) Model Problem

12 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Which relation is not a function? Regents Prep The equation x 2 + y 2 – 2x + 6y + 3 = 0 is equivalent to

13 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle Determine if the point (1, 8) lies on the locus of points that are 10 units from the point (-7,2). Model Problem

14 Course: Alg. 2 & Trig. Aim: Equation and Graph of Circle (0, 9) and (6, 1) are endpoints of a diameter of a circle. What is the equation of the circle? Model Problem


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