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

Short Answer

Expert verified
The system of equations has no solution.

Step by step solution

01

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 \[(x-2)^{2}+(y+3)^{2}=1\]
02

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.
03

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.

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