Answer:
[tex](2,\, -3)[/tex].
[tex]r = \sqrt{6}[/tex].
Step-by-step explanation:
Let [tex](a,\, b)[/tex] denote the center of this circle. Let [tex]r[/tex] ([tex]r > 0[/tex]) denote the radius of this circle. The equation of this circle would be:
[tex](x - a)^{2} + (y - b)^{2} = r^{2}[/tex].
Expand to obtain an equivalent equation:
[tex]x^{2} + (- 2\, a) + y^{2} + (- 2\, b) + (a^{2} + b^{2} - r^{2}) = 0[/tex].
Since this equation and the given equation describe the same circle, their corresponding coefficients should match.
Notice that the coefficients of [tex]x^{2}[/tex] and [tex]y^{2}[/tex] in this equation are both [tex]1[/tex]. However, the corresponding coefficients in the given equation are both [tex]2[/tex]. Thus, divide both sides of the given equation by [tex]2\![/tex] to match the coefficients of the [tex]x^{2}[/tex] and [tex]y^{2}[/tex] terms:
[tex]x^{2} + (- 4\, x) + y^{2} + 6\, y + 7 = 0[/tex].
- Coefficients of [tex]x^{2}[/tex] and [tex]y^{2}[/tex] are now matched after the division.
- Coefficients of [tex]x[/tex] should match: [tex](-4) = (-2\, a)[/tex], such that [tex]a = 2[/tex].
- Coefficients of [tex]y[/tex] should match: [tex]6 = (-2\, b)[/tex], such that [tex]b = (-3)[/tex].
The constants should also match. Thus:
[tex]a^{2} + b^{2} - r^{2} = 7[/tex].
Substitute in [tex]a = 2[/tex] and [tex]b = (-3)[/tex]; given that [tex]r > 0[/tex], the value of [tex]r[/tex] would be:
[tex]r = \sqrt{2^{2} + 3^{2} - 7} = \sqrt{6}[/tex].
