Answer:
The radius is [tex] 2\sqrt{15} [/tex]
The center is (3, 3)
Step-by-step explanation:
Equation of circle in standard form with center at (h, k) and radius r:
[tex] (x - h)^2 + (y - k)^2 = r^2 [/tex]
We need to complete the square in x and y.
[tex] x^2 + y^2 - 6x - 6y = 42 [/tex]
Separate x-terms and y-terms:
[tex] x^2 - 6x + ~~~~~y^2 - 6y + ~~~~~ = 42 + ~~~~~ + ~~~~~ [/tex]
To complete the square, you square half of the linear term's coefficient.
Linear term in x: -6x
Coefficient: -6
Half of the coefficient: -3
Square half of the coefficient: 9
We need to add 9 to both sides to complete the square in x.
The linear term in y is -6y, so we do the same and we also need to add 9 to both sides to complete the square in y.
[tex] x^2 - 6x + 9 + y^2 - 6y + 9 = 42 + 9 + 9 [/tex]
[tex] (x - 3)^2 + (y - 3)^2 = 60 [/tex]
Now that we have the equation in standard form, we can get the radius and center of the circle.
[tex] r^2 = 60 [/tex]
[tex] r = \sqrt{60} [/tex]
[tex] r = \sqrt{4 \times 15} [/tex]
[tex] r = 2\sqrt{15} [/tex]
The radius is [tex] 2\sqrt{15} [/tex]
The center is (3, 3)
