pre calculus help please, i solved them just wanted to see if there correct?

M

MoonRose

New member
Standard form of circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius

1.) This is correct, but it would be better if you just wrote it as x^2 + y^2 = 17.

2.) The center is (0, 1) and the radius is 1. To find the center, just match it up with the (h, k) values in the standard form. The square root of 1 is just 1.
 
ok write in standard form of the equation of the circle with the given characteristics:

endpoints of a diameter : ( - 4,-1) , ( 4,1)

i got
(x - 0)^2 + (y - 0)^2 = 17
is this correct....

ok second question

find the center and radius of the circle

x^2 + (y - 1)^2 = 1

this is wat i got
center is (2,0)
radius is square root of 1

is this correct
 
Standard form of circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius

1.) This is correct, but it would be better if you just wrote it as x^2 + y^2 = 17.

2.) The center is (0, 1) and the radius is 1. To find the center, just match it up with the (h, k) values in the standard form. The square root of 1 is just 1.
 
Your first answer is correct. The second is not. The equation for a circle is

(x - xc)^2 + (y - yc)^2 = r^2. where xc is the x value of the center, and yc is the y value of the center of the circle. r is the radius. So you original equation:

x^2 + (y - 1)^2 = 1 can be written (x-0)^2 + (y-1)^2 = 1^2, so you are correct that the radius is 1, but the center is (by inspection) (0,1)
 
Your first answer is correct. The second is not. The equation for a circle is

(x - xc)^2 + (y - yc)^2 = r^2. where xc is the x value of the center, and yc is the y value of the center of the circle. r is the radius. So you original equation:

x^2 + (y - 1)^2 = 1 can be written (x-0)^2 + (y-1)^2 = 1^2, so you are correct that the radius is 1, but the center is (by inspection) (0,1)
 
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