Answer:
(2.236,9.472).
Step-by-step explanation:
The standard form of a circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex] ...(1)
where, (h,k) is center and r is radius of the circle.
It is given that radius of the circle is 5 and center at (0,5). So, the equation of circle is
[tex](x-0)^2+(y-5)^2=5^2[/tex]
[tex]x^2+(y-5)^2=25[/tex] ...(2)
The given equation of line is
[tex]y=2x+5[/tex] ...(3)
We need to find the intersection point of line and circle in first quadrant.
On solving (2) and (3), we get
[tex]x^2+((2x+5)-5)^2=25[/tex]
[tex]x^2+(2x)^2=25[/tex]
[tex]x^2+4x^2=25[/tex]
[tex]5x^2=25[/tex]
[tex]x^2=5[/tex]
[tex]x=\pm \sqrt{5}=\pm 2.236[/tex]
At x=2.236,
[tex]y=2(2.236)+5=9.472[/tex]
At x=-2.236,
[tex]y=2(-2.236)+5=0.528[/tex]
It means line intersect the circle at (2.236,9.472) and (-2.236,0.528).
In first quadrant both coordinates are positive.
Therefore, the required point is (2.236,9.472).
