How do you solve this? Write the standard form of the equation of a circle with center (0, 0), given r = 8.

### 4 Answers | Add Yours

The equation of a circle with the center in M(a,b)=O(0,0) and r=8

(x-a)^2+(y-b)^2=r^2

*(x-0)^2+(y-0)^2=64*x^2+y^2=64

The standard form of a circle's equation with (-g,-f) as centre is :

x^2+y^2+2gx+2fy+c=0

The equation of the given circle with centre (0,0) and with radius 8 is (x-0)^2+(y-0)^2 =8^2.

Rearranging the equation, we get:

x^2+ y^2+2***0***x+2***0***y+**(-64)**=0, in standard form with g=0, f=0 and c=-64.

The simplified form is: x^2+y^2-64=0.

Hope this helps.

The general formula for equation of circles is:

(x-a)^2+(y-b)^2=r^2, in the standard form, where (a,b) is the centre and r is the radius.

Since centre is (0,0) and raidus=8

Sub numerals into equation

(x-0)^2 + (y-0)^2= (8)^2

**x^2+y^2=64**

the equation of the circle with centre at the origin is given by

x^2+y^2=r^2

therefore, our equation becomes

x^2+y^2=64(since r=8)

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes