Question 1

Write the standard form of the equation of the circle with the given center and radius.
-$$( 5,7 ) ; 9$$
A) $( x - 5 ) ^ { 2 } + ( y - 7 ) ^ { 2 } = 81$
B) $( x + 5 ) ^ { 2 } + ( y + 7 ) ^ { 2 } = 81$
C) $( x - 7 ) ^ { 2 } + ( y - 5 ) ^ { 2 } = 9$
D) $( x + 7 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = 9$

Accepted Answer

The standard form of the equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. Given the center $(5, 7)$ and radius $9$, the equation is $(x - 5)^2 + (y - 7)^2 = 81$, where $81$ is $9^2$.

Question 2

Write the standard form of the equation of the circle with the given center and radius.
-$( 0,0 ) ; \sqrt { 2 }$
A) $x ^ { 2 } + y ^ { 2 } = 2$
B) $x ^ { 2 } + y ^ { 2 } = \sqrt { 2 }$
C) $x ^ { 2 } + y ^ { 2 } = 1$
D) $x ^ { 2 } + y ^ { 2 } = 4$

Accepted Answer

The standard form of the equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$. For a circle centered at $(0,0)$ with radius $\sqrt{2}$, the equation is $x^2 + y^2 = (\sqrt{2})^2$, which simplifies to $x^2 + y^2 = 2$.

Question 3

Write the standard form of the equation of the circle with the given center and radius.
-$( 0,0 ) ; 5$
A) $x ^ { 2 } + y ^ { 2 } = 25$
B) $x ^ { 2 } + y ^ { 2 } = 5$
C) $x ^ { 2 } + y ^ { 2 } = 10$
D) $x ^ { 2 } - y ^ { 2 } = 5$

Accepted Answer

The standard form of the equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$. For a circle centered at $(0,0)$ with radius $5$, the equation is $x^2 + y^2 = 5^2$, which simplifies to $x^2 + y^2 = 25$.

Question 4

Complete the square and write the equation in standard form. Then give the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } - 12 x - 6 y + 33 = 0$$
A) $( x - 6 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 12$
$( 6,3 ) , r = 2 \sqrt { 3 }$
B) $( x - 6 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 12$
$( 6,3 ) , r = 12$
C) $( x - 6 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 12$
$( - 6 , - 3 ) , r = 2 \sqrt { 3 }$
D) $( x + 6 ) ^ { 2 } + ( y + 3 ) ^ { 2 } = 12$
$( - 6 , - 3 ) , \mathrm { r } = 2 \sqrt { 3 }$ Graph the equation.

Accepted Answer

To complete the square, the equation $x^2 + y^2 - 12x - 6y + 33 = 0$ is rearranged as $(x^2 - 12x) + (y^2 - 6y) = -33$. Completing the square for each binomial, we add $(\frac{12}{2})^2 = 36$ for $x$ and $(\frac{6}{2})^2 = 9$ for $y$ to both sides, resulting in $(x^2 - 12x + 36) + (y^2 - 6y + 9) = 12$, which simplifies to $(x - 6)^2 + (y - 3)^2 = 12$. The center of the circle is at $(6, 3)$, and the radius $r$ is the square root of 12, which is $2\sqrt{3}$.

Question 5

Find the center and the radius of the circle.
-$$( x + 4 ) ^ { 2 } + ( y - 9 ) ^ { 2 } = 81$$
A) $( - 4,9 ) , \mathrm { r } = 9$
B) $( 9 , - 4 ) , \mathrm { r } = 9$
C) $( 4 , - 9 ) , \mathrm { r } = 81$
D) $( - 9,4 ) , \mathrm { r } = 81$ Graph the equation and state its domain and range. Use interval notation

Accepted Answer

The equation of a circle in the form $(x - h)^2 + (y - k)^2 = r^2$ has a center at $(h, k)$ and a radius $r$. For the given equation, $(x + 4)^2 + (y - 9)^2 = 81$, the center is $(-4, 9)$ and the radius is $\sqrt{81} = 9$.

Question 6

Complete the square and write the equation in standard form. Then give the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } - 12 x - 10 y = - 12$$
A) $( x - 6 ) ^ { 2 } + ( y - 5 ) ^ { 2 } = 49$
$( 6,5 ) , r = 7$
B) $( x - 5 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 49$
$( 5,6 ) , r = 7$
C) $( x - 6 ) ^ { 2 } + ( y - 5 ) ^ { 2 } = 49$
$( - 6 , - 5 ) , r = 49$
D) $( x - 5 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 49$
$( - 5 , - 6 ) , r = 49$

Accepted Answer

To complete the square, group the x and y terms: $x^2 - 12x + y^2 - 10y = -12$. For x, add and subtract $(\frac{12}{2})^2 = 36$, and for y, add and subtract $(\frac{10}{2})^2 = 25$, resulting in $x^2 - 12x + 36 + y^2 - 10y + 25 = -12 + 36 + 25$. This simplifies to $(x - 6)^2 + (y - 5)^2 = 49$, which matches choice A. The center is at (6, 5) and the radius is $\sqrt{49} = 7$.

Question 7

Find the center and the radius of the circle.
-$x^{2}+y^{2}=36$

A)

Domain $= ( - 6,6 )$; Range $= ( - 6,6 )$
B)

Domain $= ( - \sqrt { 6 } , \sqrt { 6 } )$; Range $= ( - \sqrt { 6 } , \sqrt { 6 } )$ Graph the equation.

Accepted Answer

The answer of Find the center and the radius of...

Question 8

Write the standard form of the equation of the circle with the given center and radius.
-$( 10,2 ) ; \sqrt { 10 }$
A) $( x - 10 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 10$
B) $( x + 10 ) ^ { 2 } + ( y + 2 ) ^ { 2 } = 10$
C) $( x - 2 ) ^ { 2 } + ( y - 10 ) ^ { 2 } = 100$
D) $( x + 2 ) ^ { 2 } + ( y + 10 ) ^ { 2 } = 100$

Accepted Answer

The standard form of the equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. Given the center $(10, 2)$ and radius $\sqrt{10}$, the equation becomes $(x - 10)^2 + (y - 2)^2 = (\sqrt{10})^2$, which simplifies to $(x - 10)^2 + (y - 2)^2 = 10$.

Question 9

Write the standard form of the equation of the circle with the given center and radius.
-$( 0 , - 9 ) ; 8$
A) $x ^ { 2 } + ( y + 9 ) ^ { 2 } = 64$
B) $x ^ { 2 } + ( y - 9 ) ^ { 2 } = 8$
C) $( x + 9 ) ^ { 2 } + y ^ { 2 } = 64$
D) $( x - 9 ) ^ { 2 } + y ^ { 2 } = 64$

Accepted Answer

The standard form of the equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is its radius. Given the center $(0, -9)$ and radius $8$, the equation becomes $x^2 + (y + 9)^2 = 64$.

Question 10

Write the standard form of the equation of the circle with the given center and radius.
-$( 5,0 ) ; 10$
A) $( x - 5 ) ^ { 2 } + y ^ { 2 } = 100$
B) $( x + 5 ) ^ { 2 } + y ^ { 2 } = 100$
C) $x ^ { 2 } + ( y - 5 ) ^ { 2 } = 10$
D) $x ^ { 2 } + ( y + 5 ) ^ { 2 } = 10$

Accepted Answer

The standard form of the equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. Given the center $(5, 0)$ and radius $10$, the equation becomes $(x - 5)^2 + (y - 0)^2 = 10^2$, which simplifies to $(x - 5)^2 + y^2 = 100$.

Question 11

Complete the square and write the equation in standard form. Then give the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } + 4 x + 8 y + 16 = 0$$

A)

B)

Accepted Answer

The answer of Complete the square and write the equation...

Question 12

Find the midpoint of the line segment whose end points are given.
-$( - 5 \sqrt { 2 } , 7 \sqrt { 5 } )$ and $( - 2 \sqrt { 2 } , 10 \sqrt { 5 } )$
A) $\left( \frac { - 7 \sqrt { 2 } } { 2 } , \frac { 17 \sqrt { 5 } } { 2 } \right)$
B) $\left( \frac { 3 \sqrt { 2 } } { 2 } , \frac { 3 \sqrt { 5 } } { 2 } \right)$
C) $\left( \frac { - 3 \sqrt { 2 } } { 2 } , \frac { - 3 \sqrt { 5 } } { 2 } \right)$
D) $( - 7 \sqrt { 2 } , 17 \sqrt { 5 } )$

Accepted Answer

The midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$. Applying this formula to the given points, we get the midpoint as $\left( \frac{-5\sqrt{2} - 2\sqrt{2}}{2}, \frac{7\sqrt{5} + 10\sqrt{5}}{2} \right) = \left( \frac{-7\sqrt{2}}{2}, \frac{17\sqrt{5}}{2} \right)$.

Question 13

Complete the square and write the equation in standard form. Then give the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } - 18 x + 2 y + 82 = 4$$
A) $( x - 9 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 4$
$( 9 , - 1 ) , r = 2$
B) $( x + 1 ) ^ { 2 } + ( y - 9 ) ^ { 2 } = 4$
$( - 1,9 ) , r = 2$
C) $( x - 9 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 4$
$( - 9,1 ) , r = 4$
D) $( x + 1 ) ^ { 2 } + ( y - 9 ) ^ { 2 } = 4$
$( 1 , - 9 ) , \mathrm { r } = 4$

Accepted Answer

To complete the square, group $x$ and $y$ terms: $(x^2 - 18x) + (y^2 + 2y) = 4 - 82$. Add $(\frac{-18}{2})^2 = 81$ and $(\frac{2}{2})^2 = 1$ to both sides to maintain equality, resulting in $(x-9)^2 + (y+1)^2 = 4$. The center is $(9, -1)$ and the radius $r = \sqrt{4} = 2$.

Question 14

Find the midpoint of the line segment whose end points are given.
-$\left( - \frac { 1 } { 2 } , 1 \right)$ and $( 1,2 )$
A) $\left( \frac { 1 } { 4 } , \frac { 3 } { 2 } \right)$
B) $\left( - \frac { 3 } { 4 } , - \frac { 1 } { 2 } \right)$
C) $\left( \frac { 1 } { 2 } , 3 \right)$
D) $\left( \frac { 3 } { 4 } , \frac { 1 } { 2 } \right)$

Accepted Answer

The midpoint of a line segment is found by averaging the x-coordinates and the y-coordinates of the endpoints. For the given points $\left( - \frac { 1 } { 2 } , 1 \right)$ and $( 1,2 )$, the midpoint is $\left( \frac{ -\frac{1}{2} + 1 }{2}, \frac{1 + 2}{2} \right) = \left( \frac { 1 } { 4 } , \frac { 3 } { 2 } \right)$.

Question 15

Complete the square and write the equation in standard form. Then give the center and radius of the circle.
-$$12 x ^ { 2 } + 12 y ^ { 2 } = 144$$
A) $x ^ { 2 } + y ^ { 2 } = 12$
$( 0,0 ) , r = 2 \sqrt { 3 }$
B) $x ^ { 2 } + y ^ { 2 } = 12$
$( 0,0 ) , r = 12$
C) $x ^ { 2 } + y ^ { 2 } = 144$
$( 0,0 ) , r = 12$
D) $( \mathrm { x } - 12 ) ^ { 2 } + ( \mathrm { y } - 12 ) ^ { 2 } = 12$
$( 12,12 ) , \mathrm { r } = 2 \sqrt { 3 }$

Accepted Answer

The equation $12x^2 + 12y^2 = 144$ can be simplified by dividing each term by 12, resulting in $x^2 + y^2 = 12$. This is a circle centered at the origin $(0,0)$ with a radius $r = \sqrt{12} = 2\sqrt{3}$.

Question 16

Complete the square and write the equation in standard form. Then give the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } - 2 x - 2 y - 34 = 0$$

A)

B)

Accepted Answer

The answer of Complete the square and write the equation...

Question 17

Write the standard form of the equation of the circle with the given center and radius.
-$( 0,3 ) ; \sqrt { 13 }$
A) $x ^ { 2 } + ( y - 3 ) ^ { 2 } = 13$
B) $x ^ { 2 } + ( y + 3 ) ^ { 2 } = 13$
C) $( x - 3 ) ^ { 2 } + y ^ { 2 } = 169$
D) $( x + 3 ) ^ { 2 } + y ^ { 2 } = 169$

Accepted Answer

The standard form of the equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. Given the center $(0, 3)$ and radius $\sqrt{13}$, the equation becomes $x^2 + (y - 3)^2 = 13$.

Question 18

Find the center and the radius of the circle.
-$$( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 4$$

A)

Domain $= ( 1,5 )$, Range $= ( - 1,3 )$
B)

Domain $= ( - 5 , - 1 )$, Range $= ( - 3,1 )$

Accepted Answer

The answer of Find the center and the radius of...

Question 19

Complete the square and write the equation in standard form. Then give the center and radius of the circle.
-$$x ^ { 2 } + 8 x + 16 + y ^ { 2 } + 2 y + 1 = 36$$
A) $( x + 4 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 36$
$( - 4 , - 1 ) , r = 6$
B) $( x + 1 ) ^ { 2 } + ( y + 4 ) ^ { 2 } = 36$
$( - 1 , - 4 ) , r = 6$
C) $( x + 4 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 36$
$( 4,1 ) , r = 36$
D) $( x + 1 ) ^ { 2 } + ( y + 4 ) ^ { 2 } = 36$
$( 1,4 ) , \mathrm { r } = 36$

Accepted Answer

The given equation is already in the form of a completed square for both $x$ and $y$, with the center at $(-4, -1)$ and the radius $r = \sqrt{36} = 6$.

Quiz 2: Functions and Graphs

Write the standard form of the equation of the circle with the given center and radius. - $( 5,7 ) ; 9$

Write the standard form of the equation of the circle with the given center and radius. - $( 0,0 ) ; \sqrt { 2 }$

Write the standard form of the equation of the circle with the given center and radius. - $( 0,0 ) ; 5$

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - $x ^ { 2 } + y ^ { 2 } - 12 x - 6 y + 33 = 0$

Find the center and the radius of the circle. - $( x + 4 ) ^ { 2 } + ( y - 9 ) ^ { 2 } = 81$

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - $x ^ { 2 } + y ^ { 2 } - 12 x - 10 y = - 12$

Find the center and the radius of the circle. - $x^{2}+y^{2}=36$ $Find the center and the radius of the circle. - x^{2}+y^{2}=36 A) Domain = ( - 6,6 ) ; Range = ( - 6,6 ) B) Domain = ( - \sqrt { 6 } , \sqrt { 6 } ) ; Range = ( - \sqrt { 6 } , \sqrt { 6 } ) Graph the equation.$

Write the standard form of the equation of the circle with the given center and radius. - $( 10,2 ) ; \sqrt { 10 }$

Write the standard form of the equation of the circle with the given center and radius. - $( 0 , - 9 ) ; 8$

Write the standard form of the equation of the circle with the given center and radius. - $( 5,0 ) ; 10$

Find the midpoint of the line segment whose end points are given. - $( - 5 \sqrt { 2 } , 7 \sqrt { 5 } )$ and $( - 2 \sqrt { 2 } , 10 \sqrt { 5 } )$

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - $x ^ { 2 } + y ^ { 2 } - 18 x + 2 y + 82 = 4$

Find the midpoint of the line segment whose end points are given. - $\left( - \frac { 1 } { 2 } , 1 \right)$ and $( 1,2 )$

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - $12 x ^ { 2 } + 12 y ^ { 2 } = 144$

Write the standard form of the equation of the circle with the given center and radius. - $( 0,3 ) ; \sqrt { 13 }$

Find the center and the radius of the circle. - $( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 4$ $Find the center and the radius of the circle. - ( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 4 A) Domain = ( 1,5 ) , Range = ( - 1,3 ) B) Domain = ( - 5 , - 1 ) , Range = ( - 3,1 )$

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - $x ^ { 2 } + 8 x + 16 + y ^ { 2 } + 2 y + 1 = 36$

Quiz 2: Functions and Graphs

Write the standard form of the equation of the circle with the given center and radius. - (5,7) ;9( 5,7 ) ; 9(5,7) ;9

Write the standard form of the equation of the circle with the given center and radius. - (0,0) ;2( 0,0 ) ; \sqrt { 2 }(0,0) ;2​

Write the standard form of the equation of the circle with the given center and radius. - (0,0) ;5( 0,0 ) ; 5(0,0) ;5

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - x2+y2−12x−6y+33=0x ^ { 2 } + y ^ { 2 } - 12 x - 6 y + 33 = 0x2+y2−12x−6y+33=0

Find the center and the radius of the circle. - (x+4) 2+(y−9) 2=81( x + 4 ) ^ { 2 } + ( y - 9 ) ^ { 2 } = 81(x+4) 2+(y−9) 2=81

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - x2+y2−12x−10y=−12x ^ { 2 } + y ^ { 2 } - 12 x - 10 y = - 12x2+y2−12x−10y=−12

Find the center and the radius of the circle. - x2+y2=36x^{2}+y^{2}=36x2+y2=36

Write the standard form of the equation of the circle with the given center and radius. - (10,2) ;10( 10,2 ) ; \sqrt { 10 }(10,2) ;10​

Write the standard form of the equation of the circle with the given center and radius. - (0,−9) ;8( 0 , - 9 ) ; 8(0,−9) ;8

Write the standard form of the equation of the circle with the given center and radius. - (5,0) ;10( 5,0 ) ; 10(5,0) ;10

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - x2+y2+4x+8y+16=0x ^ { 2 } + y ^ { 2 } + 4 x + 8 y + 16 = 0x2+y2+4x+8y+16=0

Find the midpoint of the line segment whose end points are given. - (−52,75) ( - 5 \sqrt { 2 } , 7 \sqrt { 5 } ) (−52​,75​) and (−22,105) ( - 2 \sqrt { 2 } , 10 \sqrt { 5 } ) (−22​,105​)

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - x2+y2−18x+2y+82=4x ^ { 2 } + y ^ { 2 } - 18 x + 2 y + 82 = 4x2+y2−18x+2y+82=4

Find the midpoint of the line segment whose end points are given. - (−12,1) \left( - \frac { 1 } { 2 } , 1 \right) (−21​,1) and (1,2) ( 1,2 ) (1,2)

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - 12x2+12y2=14412 x ^ { 2 } + 12 y ^ { 2 } = 14412x2+12y2=144

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - x2+y2−2x−2y−34=0x ^ { 2 } + y ^ { 2 } - 2 x - 2 y - 34 = 0x2+y2−2x−2y−34=0

Write the standard form of the equation of the circle with the given center and radius. - (0,3) ;13( 0,3 ) ; \sqrt { 13 }(0,3) ;13​

Find the center and the radius of the circle. - (x−3) 2+(y−1) 2=4( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 4(x−3) 2+(y−1) 2=4

Complete the square and write the equation in standard form. Then give the center and radius of the circle. - x2+8x+16+y2+2y+1=36x ^ { 2 } + 8 x + 16 + y ^ { 2 } + 2 y + 1 = 36x2+8x+16+y2+2y+1=36