ANSWER:
(1,3) and (3, -1)
STEP-BY-STEP EXPLANATION:
The locations where they are found would be the intercepts between both functions.
The interceptions would be the solution of the system of equations proposed in the statement:
[tex]\begin{gathered} x^2+y^2=10\text{ (1)} \\ 2x+y=5\rightarrow y=5-2x\text{ (2)} \end{gathered}[/tex]We replace equation (2) in (1), and solve for x, like this:
[tex]\begin{gathered} x^2+(5-2x)^2=10 \\ x^2+25-20x+4x^2=10 \\ 5x^2-20x+25-10=0 \\ 5x^2-20x+15=0 \\ 5\cdot(x^2-4x+3)=0 \\ x^2-4x+3=0 \\ (x-1)\cdot(x-3)=0 \\ x-1=0\rightarrow x=1 \\ x-3=0\rightarrow x=3 \end{gathered}[/tex]And to calculate y, we use equation (2), like this:
[tex]\begin{gathered} y=5-2\cdot1=5-2=3\rightarrow(1,3) \\ y=5-2\cdot3=5-6=-1\rightarrow(3,-1) \end{gathered}[/tex]Which means that both are found twice at the coordinates (1,3) and (3, -1)