Question

Write the equation of the given circle. center (3, 3) radius of 2

Answer 1

(x-3)^2 + (y-3)^2 = 2^2
Hope this helped

Answer 2

General form:
`x^2+y^2-6x-6y+14=0`

Standard form:
`(x-3)^3+(y-3)^3=4`

Explanation:

We are given a circle whose center (3,3) and radius 2

Formula:

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

where, (h,k) - > (3,3)

r = 2

Substitute the value of center and r into formula

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

`x^2+9-6x+y^2+9-6y=4`

`x^2+y^2-6x-6y+14=0`

Hence, The equation of circle is
`x^2+y^2-6x-6y+14=0`