Answer:
(1.79, 7.58)
Step-by-step explanation:
Standard form equation of a circle with center (h,k) and radius r is
[tex]\displaystyle{(x-h)^2+(y-k)^2=r^2}[/tex]
Use h = 0, k = 4 and r=4 to give
--> [tex]\displaystyle{(x-0)^2+(y-4)^2=4^2}[/tex]
--> [tex]x^2 + (y-4)^2 = 16[/tex]
[tex](y-4)^2 = y^2 -8y + 16[/tex]
The line is [tex]y = 2x + 4[/tex]
Substitute for this value of y in Equation (1)
[tex]x^2 + (2x + 4 - 4)^2 = 16[/tex]
[tex]x^2 + (2x)^2 = 16[/tex]
[tex]x^2 + 4x^2 = 16[/tex]
[tex]5x^2 = 16[/tex]
[tex]x^2 = \frac{16}{5}[/tex]
[tex]x = \pm \sqrt{\frac{16}{5}}[/tex]
[tex]x = \pm \frac{4}{\sqrt{5}}[/tex]
Since we are asked to find point of intersection only on the first quadrant, we ignore the negative value of x
So [tex]x = \frac{4}{\sqrt{5} } = 1.78885 = 1.79[/tex] (rounded to 2 decimal places)
Substituting this value of x in [tex]y = 2x + 4[/tex]
[tex]y = 2(1.79) = 4 = 7.58[/tex]
So the intersection point is at
(1.79, 7.58)
See attached graph
