Question

Help guys I only have 18 minutes left!!!!!!

Answer 1

Given:

The equation of a circle is:

$(x-3)^2+y^2=32$

To find:

Center and the circumference of the circle.

Solution:

The standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$         ...(i)

Where, (h,k) is center and r is the radius.

We have,

$(x-3)^2+y^2=32$           ...(ii)

On comparing (i) and (ii), we get

$h=3,y=0,r=\sqrt{32}$

So, the center of the circle is (3,0) and the radius of the circle is $\sqrt{32}$.

Now, the circumference of the circle is:

$C=2\pi r$

$C=2(3.14)(\sqrt{32})$

$C=35.525045$

$C\approx 35.5$

Therefore, the circumference of the circle is about 35.5 units.