circles students will be able to transform an equation of a circle in standard form to center,...
TRANSCRIPT
CirclesStudents will be able to transform an equation of a circle in standard form to center, radius formby using the complete the square method.
Find the missing value to complete the square.
6. x2 – 2x + 7. x2 + 4x + 8. x2 – 6x +
Circles – Warm Up
Find the missing value to complete the square.
6. x2 – 2x + 7. x2 + 4x + 8. x2 – 6x +
Simplify.
1. 16 2. 49 3. 20 4. 48 5. 72
Solutions
6. x2 – 2x + ; c = = – = (–1)2 = 1
7. x2 + 4x + ; c = = = 22 = 4
8. x2 – 6x + ; c = = – = (–3)2 = 9b2
b2
b2
22
42
62
2 2
2 2
2 2b2b2b2
b2
1. 16 = 4
2. 49 = 7
3. 20 = 4 5 = 2 5
4. 48 = 16 3 = 4 3
5. 72 = 36 2 = 6 2
CIRCLE TERMS
EQUATION FORM
CENTER
RADIUS
MIDPOINT
FORMULA
DISTANCE
FORMULA
(x – h)² + (y – k)² = r²
(h, k )
r
2
yy,
2
xx 2121
212
212 )yy()xx(
r
C=(h , k)
Definition: A circle is an infinite number of points a set distance away from a center
Write an equation of a circle with center (3, –2) and radius 3.
Circles
(x – h)2 + (y – k)2 = r2 Use the standard form of the equation of a circle.
(x – 3)2 + (y – (–2))2 = 32 Substitute 3 for h, –2 for k, and 3 for r.
(x – 3)2 + (y + 2)2 = 9 Simplify.
Check: Solve the equation for y and enter both functions into your graphing calculator.
(x – 3)2 + (y + 2)2 = 9
(y + 2)2 = 9 – (x – 3)2
y + 2 = ± 9 – (x – 3)2
y = –2 ± 9 – (x – 3)2
Write an equation for the translation of x2 + y2 = 16 two units
right and one unit down.
Circles
(x – 2)2 + (y – (–1))2 = 16 Substitute 2 for h, –1 for k, and 16 for r 2.
(x – h)2 + (y – k)2 = r 2 Use the standard form of the equation of a circle.
(x – 2)2 + (y + 1)2 = 16 Simplify.
The equation is (x – 2)2 + (y + 1)2 = 16.
Write an equation for the translation of x2 + y2 = 16 two units
right and one unit down.
Circles
(x – h)2 + (y – k)2 = r 2 the equation of a circle.
(x – 2)2 + (y + 1)2 = 16
The equation is (x – 2)2 + (y + 1)2 = 16.Now multiply it out for standard form:
Write an equation for the translation of x2 + y2 = 16 two units
right and one unit down.
Circles
(x – h)2 + (y – k)2 = r 2 the equation of a circle.
(x – 2)2 + (y + 1)2 = 16
WRITE the equation of a circle in center, radius form
• Group the x and y terms• Move the constant term• Complete the square for x, y• Write each variable as a square
Ex: x² + y² - 4x + 8y + 11 = 0
WRITE and GRAPH• Given the equation of the circle in standard form
• x² + y² - 4x + 8y + 11 = 0• Group the x and y terms• x² - 4x + y² + 8y + 11 = 0• Complete the square for x, y• x² - 4x + 4 + y² + 8y + 16 = -11
+ 4 + 16• (x – 2)² + (y + 4)² = 9
• B) GRAPH• Plot Center (2,-4)• Radius = 3
WRITE and then GRAPH
• 4x² + 4y² + 36y + 5 = 0• Group terms, move constant, factor
out the coefficient.• Complete the square for x,y• Divide both sides by coefficient
WRITE and then GRAPH
• This time factor out the coefficients:• 4x² + 4y² + 36y + 5 = 0• Group terms, move constant, factor
out the coefficient.• 4(x²) + 4(y² + 9y + __ ) = -5 + ___• Notice that x is already done!
WRITE and GRAPH
• A) write the equation of the circle in standard form
• 4x² + 4y² + 36y + 5 = 0• Complete the square for y:• 4(x²) + 4(y² + 9y+__) = -5 +__ • 4(x²)+4(y² + 9y + 81/4) =-5+81
• 4(x²) + 4(y + 9/2)² = 76• Divide by 4:• x² + (y + 9/2)² = 19
• Center (0 , -9/2)
• Radius = 19 ≈4.5
WRITING EQUATIONSWrite the EQ of a circle that has a center of (-5,7) and passes through (7,3)
• Plot your info• Need to find values for h, k, r • (h , k) = ( , )
• How do we find r?• Use distance form. for C and P.
• Plug into circle formula• (x – h)² + (y – k)² = r²
C = (-5,7)
P = (7,3) radius
WRITING EQUATIONSWrite the EQ of a circle that has a center of (-5,7) and passes through (7,3)
• Plot your info• Need to find values for h, k, r • (h , k) = (-5 , 7)• How do we find r?• Use distance form. for C and P.
• Plug into circle formula• (x – h)² + (y – k)² = r²• (x + 5)² + (y – 7)² = ²• (x + 5)² + (y – 7)² = 160
C = (-5,7)
P = (7,3) radius
Dist = (7 −3)2 +(−5 −7)2 = 160
160
Let’s Try OneWrite the EQ of a circle that has endpoints of the diameter at (-4,2) and passes through (4,-6)
A = (-4,2)
B = (4,-6)
radius
• Plot your info• Need to find values for h, k, r • How do we find (h,k)?• Use midpoint formula
• (h , k) =• How do we find r?• Use dist form with C and B.
• Plug into formula• (x – h)² + (y – k)² = r²
Hint: Where is the center? How do you find it?
Let’s Try OneWrite the EQ of a circle that has endpoints of the diameter at (-4,2) and passes through (4,-6)
A = (-4,2)
B = (4,-6)
radius
• Plot your info• Need to find values for h, k, r • How do we find (h,k)?• Use midpoint formula
• (h , k) = (0 , -2)• How do we find r?• Use dist form with C and B.
• Plug into formula• (x – h)² + (y – k)² = r²• (x)² + (y + 2)² = 32
2
62,
2
44C
Dist = 32 =4 2
Hint: Where is the center? How do you find it?
Suppose the equation of a circle is (x – 5)² + (y + 2)² = 9• Write the equation of the new circle given that:
A) The center of the circle moved up 4 spots and left 5:
•(x – 0)² + (y – 2)² = 9 Center moved from (5,-2) (0,2)
B) The center of the circle moved down 3 spots and right 6:
•(x – 11)² + (y + 5)² = 9 Center moved from (5,-2) (11,-5)
Find the center and radius of the circle with equation
(x + 4)2 + (y – 2)2 = 36.
Let‘s Try One
The center of the circle is (–4, 2). The radius is 6.
(x – h)2 + (y – k)2 = r 2 Use the standard form.
(x + 4)2 + (y – 2)2 = 36 Write the equation.
(x – (–4))2 + (y – 2)2 = 62 Rewrite the equation in
standard form.
h = –4 k = 2 r = 6 Find h, k, and r.
Graph (x – 3)2 + (y + 1)2 = 4.
Let’s Try One
(x – h)2 + (y – k)2 = r 2 Find the center and radius of the circle.
(x – 3)2 + (y – (–1))2 = 4
h = 3 k = –1 r 2 = 4, or r = 2
Draw the center (3, –1) and radius 2.Draw a smooth curve.
Solving linear quadratic systemsDo now: page 182 # 37
Solving a circle-line system
Y=-x-2
• Plot the line b=-2 m = -1• Plot the center and count in four
directions for the radius• Find the points of intersection
• y=-x -2• x² + (y + 2)² = 32
Solving a circle-line system
(-4,2)
(4,-6)
Y=-x-2
• Plot the line b=-2 m = -1• Plot the center and count in four
directions for the radius• Find the points of intersection
• y=-x -2• x² + (y + 2)² = 32
c=(0,−2), r ≈5.7
Solving a circle-line system
By algebraic method: substitute –x-2 for y into second equation
• y=-x -2• x² + (y + 2)² = 32
Solving a circle-line system
By algebraic method: substitute –x-2 for y into second equation
• y=-x -2• x² + (y + 2)² = 32• x² + (-x - 2+ 2)² = 32• x² + ( -x)² = 32• 2x2=32• X2=16• X=4, x=-4 solve for y:• Y=-4-2 y=4-2• Y=-6 (4,-6) y=2 (-4, 2)
Lets solve this one both ways
(x−2)2 +(y+2)2 =9x+ y=−3
Lets solve this one both ways
(x−2)2 +(y+2)2 =9x+ y=−3
(−1,−2) and (2,−5)