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Aim: How do we find the equation of the locus of points at a given distant from a given point?. 4. 4. 4. 4. Do Now: What is the locus of points 4 units from a given point?. LOCUS. Fundamental Locus 5. 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? 4 4 4 4 Do Now: What is the locus of points 4 units from a given point? LOCUS Fundamental Locus 5 The locus is a circle whose center is the given point and that has a radius of 4 units.
3 (0,0) 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. Fundamental Loci -5 Circle Basics Describe the locus of points 3 units from the from the origin. The locus is a circle with a radius of 3 units and whose center is the origin What is the equation of this locus?
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,-5) (-3,4) (4,3) (-4,3) 5 (-5,0) (5,0) (0,0) (-4,-3) (-3,-4) (4,-3) (3,-4) (0,-5) P(x,y) P(3,4) 32 + 42 = 52 = 25 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
Model Problem 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 r = 5 x2 + y2 = 25
The locus is a circle whose center is origin and whose radius is the square root of 45 or 4 x2 + y2 = 100 Describe fully the locus of points for the given equation. A) x2 + y2 = 45 B) 4x2 + 4y2 = 64 x2 + y2 = 16 The locus is a circle whose center is origin and whose radius of 4. Is (8,6) located on the locus of a points whose distance from the origin is 10? 82 + 62 = 100 YES 64 + 36 = 100
Model Problem 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 the origin and with a radius of 4 units. x2 + y2 = 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 x2 + y2 = 36.
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 (4,3) (4, 3) (x,y) = 5 (x – h)2 + (y – k)2 = r2 (h,k) What is the value of (h, k)? What is the equation for this circle whose radius is 5 with center at (4, 3)? The standard form of an equation of a circle with center (h, k) and radius r is (x – h)2 + (y – k)2 = r2 (x – 4)2 + (y – 3)2 = 52 (x – 4)2 + (y – 3)2 = 25
= 5 r O’ (h,k) 3 (4,3) 4 O (0,0) x2 + y2 = 25 (x–4)2 + (y–3)2 = 52 is a translation of x2 + y2 = 25 under the rule T4,3(x,y) = (x + 4, y + 3), or Circle with Center (h,k) What is the equation for the circle whose radius is 5 with center at (4, 3)? (x–4)2 + (y–3)2 = 52 The standard form of an equation of a circle with center (h, k) and radius r is (x – h)2 + (y – k)2 = r2
Model Problem 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 Origin -(0,0) YES (0 + 5)2 + 02 = 25 (5)2 = 25
Model Problem 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 complete the squares x2 + 14x + ( )+ y2 + 2y + ( )= -40 49 1 + 49 + 1 rewrite as squares of binomials (x + 7)2 + (y + 1)2 = 10 center: (h, k) = (-7, -1)
Regents Prep Which relation is not a function? The equation x2 + y2 – 2x + 6y + 3 = 0 is equivalent to
Model Problem Determine if the point (1, 8) lies on the locus of points that are 10 units from the point (-7,2).
Model Problem (0, 9) and (6, 1) are endpoints of a diameter of a circle. What is the equation of the circle?