Answer:
A. [tex]C(x,y) = (3,-1)[/tex], [tex]r = 2[/tex], x-Intercepts: [tex]x = 3\pm\sqrt{3}[/tex], y-Intercepts: none.
Step-by-step explanation:
Let [tex]x^{2}+y^{2}-6\cdot x + 2\cdot y + 6 = 0[/tex] be the general equation of the circle, we must transform the expression into standard form to determine its center, radius and intercepts. The procedure is shown below:
1) [tex]x^{2}+y^{2}-6\cdot x + 2\cdot y + 6 = 0[/tex] Given.
2) [tex](x^{2}-6\cdot x)+(y^{2}+2\cdot y) +6= 0[/tex] Commutative and associative properties.
3) [tex](x^{2}-6\cdot x)+(y^{2}+2\cdot y) + 6 + 3 + 1 = 3 + 1[/tex] Compatibility with addition.
4) [tex](x^{2}-6\cdot x +9)+(y^{2}+2\cdot y +1) = 4[/tex] Definition of addition/Commutative and associative properties.
5) [tex](x-3)^{2} + (y+1)^{2} = 4[/tex] Perfect square trinomial/Result.
The equation of the circle centered in (h, k) in standard form is defined as:
[tex](x -h)^{2} + (y-k)^{2} = r^{2}[/tex] (Eq. 1)
Where:
[tex]h[/tex], [tex]k[/tex] - Coordinate of the center of the circle, dimensionless.
[tex]r[/tex] - Radius of the circle, dimensionless.
By direct comparison we find that circle is centered in [tex]C(x, y) = (3, -1)[/tex] and has a radius of 2.
Finally, we obtain the intercepts of the given function:
x-Intercepts ([tex]y = 0[/tex])
[tex](x-3)^{2} + (0+1)^{2} = 4[/tex]
[tex]x^{2}-6\cdot x +9+1 = 4[/tex]
[tex]x^{2}-6\cdot x +6=0[/tex]
Roots are found analitically by Quadratic Formula:
[tex]x = 3\pm\sqrt{3}[/tex]
y-Intercepts ([tex]x=0[/tex])
[tex](0-3)^{2}+(y+1)^{2} = 4[/tex]
[tex]9+y^{2}+2\cdot y +1 = 4[/tex]
[tex]y^{2}+2\cdot y +6 = 0[/tex]
Roots are found analitically by Quadratic Formula:
[tex]y = -1\pm i\,\sqrt{5}[/tex]
In a nutshell, there are no y-Intercepts.
We include a graphic including circle, center and x-Intercepts.
Finally, we came to the conclusion that correct answer is A.
