Recall the equation for a circle with center $(h, k)$ and radius $r$. At what point in the first quadrant does the line with equation $y=0.5 x+4$ intersect the circle with radius 6 and center $(0,4)$ ? \[ \begin{array}{l} x= \\ y= \end{array} \] Enter your answer correct to 3 decimal places .

Step 1 ：The equation for a circle with center \((h, k)\) and radius \(r\) is given by \((x-h)^2 + (y-k)^2 = r^2\).

Step 2 ：Given that the circle has a center at \((0,4)\) and a radius of 6, the equation of the circle is \((x-0)^2 + (y-4)^2 = 6^2\), which simplifies to \(x^2 + (y-4)^2 = 36\).

Step 3 ：The equation of the line is \(y=0.5x+4\).

Step 4 ：Substituting \(y\) into the equation of the circle gives \(x^2 + (0.5x)^2 = 36\), which simplifies to \(x^2 + 0.25x^2 = 36\).

Step 5 ：Combining like terms gives \(1.25x^2 = 36\), and solving for \(x\) gives \(x^2 = 36/1.25\).

Step 6 ：Taking the square root of both sides gives \(x = \sqrt{36/1.25}\).

Step 7 ：Since we are looking for the point in the first quadrant, we take the positive root, so \(x = \sqrt{36/1.25} \approx 5.366\).

Step 8 ：Substituting \(x\) into the equation of the line gives \(y = 0.5(5.366) + 4 \approx 6.683\).

Step 9 ：So, the point of intersection in the first quadrant is approximately \((5.366, 6.683)\).

Step 10 ：Checking the solution, we substitute \(x\) and \(y\) into the equation of the circle and verify that the left-hand side equals the right-hand side, confirming that the solution is correct.

Step 11 ：The final answer is \(\boxed{(5.366, 6.683)}\).