(a) Suppose you want to solve the quartic equation [math]x^4+ax^2+bx+c=0[/math] by intersecting the parabola [math]y=x^2[/math] with a circle, so that the x-coordinates of the intersections are the roots of the equation. What are the center and the radius of the circle you should draw?[br](b) Let d > 0. Suppose you want to solve the quartic equation [math]x^4+ax^3+bx+cx+d=0[/math] by intersecting the hyperbola xy = [math]\sqrt{d}[/math] with a circle so that the x-coordinates of the intersections are the roots of the equation. What are the center and the radius of the circle you should draw?

(a) We are given [math]y=x^2[/math]. We know the equation for a circle is [math]\left(x-h\right)^2+\left(y-k\right)^2=r^2[/math]. By foiling we have: [math]x^2-2hx+h^2+x^4-2kx^2+k^2=r^2[/math]. After factoring and moving [math]r^2[/math] to the other side of the equation we have: [math]x^4+\left(1-2k\right)x^2-2hx+h^2+k^2-r^2=0[/math]. Thus, [math]a=1-2k[/math] and [math]b=-2h[/math]. When substituting and solving for h we have[math]h=-\frac{b}{2}[/math] and [math]k=\frac{\left(-a+1\right)}{2}[/math] . We can see that [math]c=h^2+k^2-r^2[/math]. When replacing h and k we solve for c and get [math]c=\left(-2h\right)^2+\left(\frac{\left(1-a\right)}{2}\right)^2-r^2[/math]. Thus, [math]r^2=\left(-2h\right)^2+\left(\frac{\left(1-a\right)}{2}\right)^2-c[/math]and [math]r^{2=}4h^2+\left(\frac{\left(1-a\right)}{2}\right)^2-c[/math]. And [math]r=\sqrt{\frac{\left(b^2\right)}{4}+\left(\frac{\left(1-a\right)}{2}\right)^2-c}[/math]. This is then the radius of the circle with center c. In GeoGebra we can first graph [math]y=x^2[/math]. Then create the circle. Then, graph the equation [math]x^4+ax^2+bx+c=0[/math]. Next we can construct the intersection points of the equation [math]y=x^2[/math] and the circle. You can notice that the x values of the roots are equivalent to the x values of the intersection points. [br](b) We can start by graphing the quartic equation given: [math]x^4+ax^3+bx^2+cx+d[/math]. Next we can graph the equation [math]y=\frac{\sqrt{d}}{x}[/math]This time we will be constructing a circle with center ([math]\frac{\left(-a\right)}{2},\frac{\left(-c\right)}{2\sqrt{d}}[/math]) and radius [math]r^2=\left(-2h\right)^2+\left(\frac{\left(1-a\right)}{2}\right)^2-c[/math]. Construct the intersection points of the equation [math]y=\frac{\sqrt{d}}{x}[/math]and the circle. As you can see again, the x coordinates of the roots and the intersection points are equivalent.