Answer:
General form: [tex]x^2+y^2-6x-6y+14=0[/tex]
Standard form: [tex](x-3)^3+(y-3)^3=4[/tex]
Step-by-step explanation:
We are given a circle whose center (3,3) and radius 2
Formula:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
where, (h,k) - > (3,3)
r = 2
Substitute the value of center and r into formula
[tex](x-3)^3+(y-3)^3=2^2[/tex]
[tex]x^2+9-6x+y^2+9-6y=4[/tex]
[tex]x^2+y^2-6x-6y+14=0[/tex]
Hence, The equation of circle is [tex]x^2+y^2-6x-6y+14=0[/tex]
