x^2+y^2-4x-2y-4=0 and x^2+y^2-12x-8y-36=0
You asked:
Solve the system of equations:\[\begin{aligned}{x}^{2} + {y}^{2} - 4 x - 2 y - 4 &= 0\\{x}^{2} + {y}^{2} - 12 x - 8 y - 36 &= 0\end{aligned}\]
MathBot Answer:
The system of equations has \(2\) solutions.
\[x = - \frac{58}{25} - \frac{18 \sqrt{14} i}{25}, y = - \frac{56}{25} + \frac{24 \sqrt{14} i}{25}\]\[x = - \frac{58}{25} + \frac{18 \sqrt{14} i}{25}, y = - \frac{56}{25} - \frac{24 \sqrt{14} i}{25}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).