ANSWER and EXPLANATION
1. We want to find the center of the given equation:
[tex]x^2+y^2-6x+4y=12[/tex]To do this, complete the square for the two variables in the equation, x and y:
[tex]\begin{gathered} x^2-6x+y^2+4y=12 \\ \\ x^2-6x+(-\frac{6}{2})^2+y^2+4y+(\frac{4}{2})^2=12+(-\frac{6}{2})^2+(\frac{4}{2})^2 \\ \\ x^2-6x+9+y^2+4y+4=25 \end{gathered}[/tex]Now, factorize the x and y parts of the equation and write the equation of the circle in general form:
[tex]\begin{gathered} (x^2-3x-3x+9)+(y^2+2y+2y+4)=25 \\ \\ x(x-3)-3(x-3)+y(y+2)+2(y+2)=5^2 \\ \\ (x-3)^2+(y+2)^2=5^2 \end{gathered}[/tex]In the general form of the equation of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]the center of the circle is (h, k).
Therefore, comparing the given equation to the general equation of a circle, the center of the circle is:
[tex](3,-2)[/tex]2. Now, we want to graph the circle. The graph of the circle is given below:
3. To find the x and y-intercepts of the circle, we have to find the points where the graph of the circle touches the x and y axes.
For the y-intercepts:
[tex](0,-6)\text{ and }(0,2)[/tex]That is the answer.
Since we want to write the x-intercepts in radical form, we can solve for them from the given equation.
To find the x-intercepts, solve for x when y is equal to 0:
[tex]\begin{gathered} x^2+0^2-6x+(0)y=12 \\ \\ x^2-6x-12=0 \end{gathered}[/tex]Solve using the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]where a = 1, b = -6, c = -12
Therefore, the x-intercepts are:
[tex]\begin{gathered} x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(-12)}}{2(1)}=\frac{6\pm\sqrt{36+48}}{2} \\ \\ x=\frac{6\pm\sqrt{84}}{2}=\frac{6\pm2\sqrt{21}}{2} \\ \\ x=3+\sqrt{21}\text{ and }x=3-\sqrt{21} \end{gathered}[/tex]Those are the x-intercepts in radical form.
