The Solution:
The equation for a circle with center ( h , k ) and radius r is:
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ Given: \\ Center=(h,k)=(0,5)\Rightarrow h=0,k=5 \\ radius=r=3 \end{gathered}[/tex]Required:
To find the point in the first quadrant, where the line y = x + 5 intersects the given circle.
Step 1:
Write out the equation of the circle and that of the line.
[tex]\begin{gathered} (x-0)^2+(y-5)^2=3^2 \\ x^2+(y-5)^2=9 \\ x^2+(y-5)^2-9=0\text{ \lparen equation of the circle\rparen} \\ y=x+5\text{ \lparen equation of a line\rparen} \end{gathered}[/tex]Plotting the graphs of both equations, we have:
Therefore, the correct answer is:
[tex]\begin{gathered} (2.121,7.121) \\ x=2.121 \\ y=7.121 \end{gathered}[/tex]