# 8_1

- Author:
- angela

## 8_1a GeoGebra

## 8_1b

## 8_1

(a) Suppose you want to solve the quartic equation by intersecting the parabola 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?
(b) Let d > 0. Suppose you want to solve the quartic equation by intersecting the hyperbola xy = 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?

## Solution:

(a) We are given . We know the equation for a circle is . By foiling we have: . After factoring and moving to the other side of the equation we have: . Thus, and . When substituting and solving for h we have and . We can see that . When replacing h and k we solve for c and get . Thus, and . And . This is then the radius of the circle with center c. In GeoGebra we can first graph . Then create the circle. Then, graph the equation . Next we can construct the intersection points of the equation and the circle. You can notice that the x values of the roots are equivalent to the x values of the intersection points.
(b) We can start by graphing the quartic equation given: . Next we can graph the equation This time we will be constructing a circle with center ( ) and radius . Construct the intersection points of the equation and the circle. As you can see again, the x coordinates of the roots and the intersection points are equivalent.