Aim: How do we find the equation of the locus of points at a given distant from a given point?
description
Transcript of Aim: How do we find the equation of the locus of points at a given distant from a given point?
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.
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Aim: How do we find the equation of the locus of points at a given distant from a given point?
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)
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What is the equation of this locus?
Circle Basics
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)(x1 x2 )2 ( y1 y2 )2 d
(x1 0)2 ( y1 0)2 5
(x)2 (y)2 5
x2 y2 25
(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
x2 + y2 = r2
32 + 42 = 52 = 25
Course: Alg. 2 & Trig.Aim: Equation and Graph of Circle
The equation x2 + y2 = 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
x2 + y2 = r2
r2 = 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.
x2 + y2 = r2
x2 + y2 = 25r = 5
Model Problem
Course: Alg. 2 & Trig.Aim: Equation and Graph of Circle
Describe fully the locus of points for the given equation.
A) x2 + y2 = 45
B) 4x2 + 4y2 = 64
The locus is a circle whose center is origin and whose radius is the square root of 45 or 3 5
The locus is a circle whose center is origin and whose radius of 4.
4x2 + y2 = 16
Is (8,6) located on the locus of a points whose distance from the origin is 10?
x2 + y2 = 100 82 + 62 = 100
64 + 36 = 100 YES
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 x2 + y2 = 16 and x2 + y2 = 64.
x2 + y2 = 16 is a circle with center at theorigin and with a radius of 4 units.
x2 + y2 = 64 is a circle with center at theorigin 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 x2 + y2 = 36.
Model Problem
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)
(x,y)(x h)2 (y k)2 r
(x – h)2 + (y – k)2 = r2
What is the value of (h, k)? (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 = 52
= 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 = r2
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 = r2
What is the equation for the circle whose radius is 5 with center at (4, 3)?
(x – 4)2 + (y – 3)2 = 52
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r = 5
(4,3)(h,k)
O’
x2 + y2 = 25
(0,0)O
(x – 4)2 + (y – 3)2 = 52 is a translation of x2 + y2 = 25under the rule T4,3(x,y) = (x + 4, y + 3), or
OT4 ,3 O'
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 = r2
h = -3, k = 5, r = 6
(x – -3)2 + (y – 5)2 = 62
(x + 3)2 + (y – 5)2 = 36
Find the center and radius of the circle whose equation is (x + 5)2 + y2 = 25.
Does the origin lie on the locus of the points represented by the equation?
Center - (-5,0) Radius = 5
(0 + 5)2 + 02 = 25Origin -(0,0)
(5)2 = 25 YES
Model Problem
Course: Alg. 2 & Trig.Aim: Equation and Graph of Circle
Find the coordinates of the center of the circle whose equation is
x2 + 14x + y2 + 2y = -40
Standard equation of circle(x – h)2 + (y – k)2 = r2
x2 + 14x + ( )+ y2 + 2y + ( )= -40
complete the squares
49 1 + 49 + 1
(x + 7)2 + (y + 1)2 = 10
rewrite as squares of binomials
center: (h, k) = (-7, -1)
Model Problem
Course: Alg. 2 & Trig.Aim: Equation and Graph of Circle
Which relation is not a function?
Regents Prep
2 2
2
1) 2 4 2) 4
3) 4 4 4) 4
x y x y
x x y xy
The equation x2 + y2 – 2x + 6y + 3 = 0 is equivalent to
2 2
2 2
2 2
2 2
1) 1 3 3
2) 1 3 7
3) 1 3 7
4) 1 3 10
x y
x y
x y
x y
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
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