Answer:
Step-by-step explanation:
[tex]x+y=4[/tex]
[tex](x+3)^{2}+(y-2)^{2}=10[/tex]
From first equation
[tex]x=4-y[/tex]
Substituting in the second equation
[tex](4-y+3)^{2}+(y-2)^{2}=10\\\Rightarrow (7-y)^2+y^2+4-4y=10\\\Rightarrow 49+y^2-14y+y^2+4-4y=10\\\Rightarrow 2y^2-18y+43=0[/tex]
Solving the equation we get
[tex]y=\frac{-\left(-18\right)+\sqrt{\left(-18\right)^2-4\cdot \:2\cdot \:43}}{2\cdot \:2}, \frac{-\left(-18\right)-\sqrt{\left(-18\right)^2-4\cdot \:2\cdot \:43}}{2\cdot \:2}\\\Rightarrow y=\frac{9}{2}+i\frac{\sqrt{5}}{2}, \frac{9}{2}-i\frac{\sqrt{5}}{2}[/tex]
Hence the circle and the line will not intersect and have no solutions
