Problem 38 - College Algebra

Question

Solve each system by the method of your choice. $$\left\\{\begin{array}{l} x^{2}-y^{2}-4 x+6 y-4=0 \\ x^{2}+y^{2}-4 x-6 y+12=0 \end{array}\right.$$

Step-by-Step Solution

Answer

The system of equations has no solution.

Step 1: Reformulate Equations

Reformulate the given equations to match the general form \((x-a)^{2}+(y-b)^{2}=r^{2}\). The equations become: \[(x-2)^{2}+(y-3)^{2}=9\] and \[(x-2)^{2}+(y+3)^{2}=1\]

Step 2: Find Intersection Points

Set the two equations equal to each other to find the intersection points. Plotting to help visualize, we find \(x-2\) and \(y-3\), or \(y+3\) matches in the two circle equations, leaving us with two possibilities: \[9=1\] and \[-9=1\] Both possibilities are not true, which indicates no solution.

Step 3: Conclude

Since the two possibilities yielded false statements, it is concluded that the two circles represented by the original system of equations do not intersect. Therefore, the system of equations has no solution.