(x- h)^2+(y- k)^2= r^2Here, the center is the point ( h, k) and the radius is r. We will rewrite the given equation to match this form, and then we can identify the center and the radius. In this case we only need to subtract zero from the variable terms.
x^2+y^2 = 4.5^2
⇕
(x- 0)^2+(y- 0)^2= 4.5^2
The center of the circle is the point ( 0, 0), and its radius is 4.5.
Finally, we can graph the circle using this information.
b Like in Part A, let's start by recalling the standard equation of a circle.
(x- h)^2+(y- k)^2= r^2Here, the center is the point ( h, k) and the radius is r. We will rewrite the given equation to match this form, and then we can identify the center and the radius. Like in Part A, we need to subtract zero from the variable terms, but unlike in Part A, we also need to change the constant term into a square.
x^2+y^2 = 75
⇕
(x- 0)^2+(y- 0)^2=( sqrt(75))^2
The center of the circle is the point ( 0, 0), and its radius is sqrt(75).
Finally, we can graph the circle using this information.
c Like in Parts A and B, let's start by recalling the standard equation of a circle.
(x- h)^2+(y- k)^2= r^2Here, the center is the point ( h, k) and the radius is r. We will rewrite the given equation to match this form, and then we can identify the center and the radius. This time we need to subtract zero from the y-variable and write the constant term as a square of a non-negative number.
(x-3)^2+y^2 = 1
⇕
(x- 3)^2+(y- 0)^2= 1^2
The center of the circle is the point ( 3, 0), and its radius is 1.
Finally, we can graph the circle using this information.
d Like in Parts A-C, let's start by recalling the standard equation of a circle.
(x- h)^2+(y- k)^2= r^2Here, the center is the point ( h, k) and the radius is r. We will rewrite the given equation to match this form, and then we can identify the center and the radius. This time we complete the square the square twice — once per each variable.
