e Repeat the process from Parts A-D for the given equation.
A
a 9
B
b 4
C
c (x+3)^2+(y-2)^2=9
D
dCenter: (-3,2) Radius: 3
E
eCenter: (-2,10) Radius: 2
Practice makes perfect
a To determine the number c we will use completing the square. Let's look at the given equation.
x^2+6x+y^2-4y= - 4
In a quadratic expression b is the linear coefficient. For the given equation, we have that b= 6 for the x-terms. Let's first calculate ( b2 )^2 for the x-terms, as this will be the number c we need to add to each side of the equation.
b For the given equation we have b= -4 for the y-terms. Let's repeat the process from Part A to find ( b2 )^2. This will represent the number d we need to add to each side of the equation so that y^2-4y+d can be factored into a perfect square trinomial.
c To rewrite the given equation we will add ( b2 )^2= 9 and ( b2 )^2=4 to both sides of our equation. Then, we will factor the two trinomials on the left-hand side.
d The standard form for the equation of a circle with center (h,k) and radius r is (x-h)^2+(y-k)^2=r^2. To find the center and radius, let's look at our equation in rewritten form from Part C.
(x+3)^2+(y-2)^2=9
⇕
(x-( -3))^2+(y- 2)^2=(sqrt(9))^2
We see above that h= -3, k= 2, and r=sqrt(9)=3. Therefore, the center is at ( -3, 2). The radius is 3 units.
e We will repeat the steps from Parts A-D to find the center and radius of the given equation. We will start by moving the constant to the left-hand side of the equation by subtracting 100 from both sides.
x^2+4x+y^2-20y+100=0
⇕
x^2+4x+y^2-20y=- 100
Now we need determine the numbers c and d we will add to both sides of the equation. For the given equation, we have that b= 4 for the x-terms and b= -20 for the y-terms. Let's first calculate ( b2 )^2 for the x-terms, as this will be the number c we need to add to each side of the equation.
For the y-terms, ( b2 )^2=100.
To rewrite the equation, we will add ( b2 )^2= 4 and ( b2 )^2=100 to both sides of our equation. Then, we will factor the two trinomials on the left-hand side of the equation.
Now that we have our equation written in standard form for the equation of a circle, we can use it to find the center and radius.
(x+2)^2+(y-10)^2=4
⇕
(x-( -2))^2+(y- 10)^2=(sqrt(4))^2
We see above that h= -2, k= 10, and r=sqrt(4)=2. Therefore, the center is at ( -2, 10). The radius is 2 units.
