write the equation of the circle in standard form and then identify its center and radius?

general equation of a circle in standard form:

(x - h)^2 + (y - k)^2 = r^2

general equation of a circle of radius r and center (0,0)

x^2 + y^2 = r^2

since you're given (4/3)X^2 + (4/3)Y^2 = 1

then the standard form will be (x - 0)^2 + (y - 0)^2 = 3/4

the center is (0,0) and the radius is sqrt(3/4)

hope this helps!

Well the standard form of a circle is

(x - h)^2 + (y - k)^2 = r^2

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

Taking your equation if we multiply both sides by 3/4 we get

x^2 + y^2 = 3/4

This is the same thing as writing

(x - 0)^2 + (y - 0)^2 = 3/4

Now hopefully you see

h = 0, k = 0 and r^2 = 3/4 ----> r = sqrt(3/4)

So your center is at (0,0) and it has radius of sqrt(3/4)

Multiply both sides by 3/4, then the given equation becomes

X^2 + Y^2 = 3/4

and the center of the circle is at (0,0) and with radius of (1/2)sqrt3.

Hope this helps.

4/3X²+4/3y²=1

X²+Y²=3/4

the center is in (0,0) and the radius is (√3)/2

For the equation you've given, the centre is the origin and the radius is (3/4)^1/2

(3 / 4)(4 / 3)x^2 + (3 / 4)(4 / 3)y^2 = (3 / 4)(1)

x^2 + y^2 = 3 / 4

centre (0, 0) and radius = sqrt(3) / 2

(4/3)X² + (4/3)Y² = 1

(X-0)² + (Y-0)² = 3/4

center (0,0)

radius √3/2

x^2 + y^2 = 3/4

centre is(0,0); radius = √3/2

x^2+y^2 = 3/4

center (0,0)

radius √(3)/2

agree with moise.