a. The equation in center-radius form is already provided. The center-radius form of a circle equation is given by:
[tex]\[ (x-h)^2 + (y-k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
In this case, the equation is:
[tex]\[ (x-3)^2 + (y-3)^2 = 16 \][/tex]
This represents a circle with center [tex]\((3, 3)\)[/tex] and radius [tex]\(4\)[/tex] (since [tex]\( r^2 = 16 \)[/tex], thus [tex]\( r = \sqrt{16} = 4 \)[/tex]).
b. To convert the equation from center-radius form to general form, we need to expand it and then combine like terms.
Starting with the equation:
[tex]\[ (x-3)^2 + (y-3)^2 = 16 \][/tex]
Expanding each squared term:
[tex]\[ (x-3)^2 = x^2 - 6x + 9 \][/tex]
[tex]\[ (y-3)^2 = y^2 - 6y + 9 \][/tex]
Substitute these back into the original equation:
[tex]\[ x^2 - 6x + 9 + y^2 - 6y + 9 = 16 \][/tex]
Combine like terms:
[tex]\[ x^2 + y^2 - 6x - 6y + 18 = 16 \][/tex]
Subtract 16 from both sides to set the equation to zero (standard form of a circle equation):
[tex]\[ x^2 + y^2 - 6x - 6y + 2 = 0 \][/tex]
Thus, the equation in general form is:
[tex]\[ x^2 + y^2 - 6x - 6y + 2 = 0 \][/tex]
