Find the center and raduis of the circle. 3(x-3)^2+3y^2=3

Question

Can you please show me each step.

Thank you!!

Answer 1

Given equation is 3(x-3)^2+3y^2 = 3

Divide to each side by 3.

[3(x-3)^2]/3+(3y^2)/3 = 3/3

(x-3)^2+y^2 = 1

(x-3)^2+(y-0)^2 = 1^2

Compare it to circle standard form (x-h)^2+(y-k)^2 = r^2

h = 3, k = 0, r = 1

Radious is 1, center is (3,1).

Comments

(x - 3)² + (y - 0)² = 1² .

Compare the above equation with the standard form of a circle equation : ( x - h )² + ( y - k )² = r², where, (h, k) is the center of the circle, and r is the radius.

The center (h, k) = (3, 0).

The radius (r) = 1 unit.