Question:
Solution:
An equation of the circle with center (h,k) and radius r is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]This is called the center-radius form of the circle equation.
Now, in this case, notice that the center of the circle is (h,k) = (-3,3) and its radius is r = 3 so that the center-radius form of the circle would be:
[tex](x+3)^2+(y-3)^2=3^2[/tex]To obtain the general form, we must solve the squares of the previous equation:
[tex](x+3)^2+(y-3)^2-3^2\text{ = 0}[/tex]this is equivalent to:
[tex](x^2+6x+3^2)+(y^2-6y+3^2)\text{ - 9 = 0}[/tex]this is equivalent to
[tex]x^2+6x+9+y^2-6y\text{ = 0}[/tex]this is equivalent to:
[tex]x^2+y^2+6x-6y\text{ +9= 0}[/tex]so that, the general form equation of the circle would be:
[tex]x^2+y^2+6x-6y\text{ +9= 0}[/tex]thus, the correct answer is:
CENTER - RADIUS FORM:
[tex](x+3)^2+(y-3)^2=3^2[/tex]GENERAL FORM:
[tex]x^2+y^2+6x-6y\text{ +9= 0}[/tex]
