(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 identify the center and the radius.
(y-7)^2=25-(x-3)^2
In this case, we are already given two squared binomials. In order to obtain the standard equation we should add (x-3)^2 to both sides of the equation.
We obtained the standard equation of the circle, so we can identify its center and radius. The center of the circle is ( 3, 7) and its radius is 5.
b
Examining the given circle, we can see that there are two terms containing y-variable.
x^2+ y^2+10 y=-9
To write the standard equation of the circle, we will need to complete the square for y. In this case, we should add 5^2 to both sides of the equation.
From the standard equation of the given circle, we can see that its center is ( -9, 4) and its radius is 5sqrt(2).
d
Similarly as in Part A, the squares are already given. However, let's rearrange the sum on the left-hand side by using Commutative Property of Addition to obtain the standard equation of the circle.
y^2+(x-3)^2=1
⇕
(x- 3)^2+(y- 0)^2= 1^2
The center of the given circle is ( 3, 0) and its radius is 1.
