Answer:
2 square root 5
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
The equation of the circle is equal to
[tex](x-h)^2+(y-k)^2=r^2[/tex]
where
(h,k) is the center and r is the radius of the circle
In this problem we have
center at (0,0) and radius equal 6 units
so
[tex]x^2+y^2=6^2\\x^2+y^2=36[/tex]
Remember that
If a point lie on the circle, then the point must satisfy the equation of the circle
Substitute the x-coordinate of point Q in the equation of the circle and solve for the y-coordinate of point Q
For x=4
[tex]4^2+y^2=36\\y^2=36-16\\y^2=20[/tex]
[tex]y=\pm\sqrt{20}[/tex]
simplify
[tex]y=\pm2\sqrt{5}[/tex]
The point Q lie on the first Quadrant (see the picture)
The y-coordinate is positive
therefore
The y-coordinate of point Q is [tex]2\sqrt{5}[/tex]
