Answer:
[tex]\displaystyle \Big(\frac{3\sqrt{5}}{5},\frac{6\sqrt5}{5}+4\Big)[/tex]
Step-by-step explanation:
Recall that the equation for a circle is given by:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Where (h, k) is the center and r is the radius.
Then for a center of (0, 4) and a radius of 3, our equation is:
[tex]x^2+(y-4)^2=9[/tex]
We want to know at what point does the circle intersect with the line:
[tex]y=2x+4[/tex]
Therefore, we can solve for x. To do so, substitute the linear equation into the circle equation:
[tex]x^2+((2x+4)-4)^2=9[/tex]
Simplify:
[tex]x^2+(2x)^2=9[/tex]
Square:
[tex]x^2+4x^2=5x^2=9[/tex]
Divide both sides by 5:
[tex]\displaystyle x^2=\frac{9}{5}[/tex]
Therefore:
[tex]\displaystyle x=\pm\frac{3}{\sqrt{5}}=\pm\frac{3\sqrt{5}}{5}[/tex]
In QI, x is always positive, so we only need to consider the positive case:
[tex]\displaystyle x=\frac{3\sqrt{5}}{5}[/tex]
Using the linear equation again, we can see that:
[tex]\displaystyle y=2\Big(\frac{3\sqrt{5}}{5}\Big)+4=\frac{6\sqrt{5}}{5}+4[/tex]
Therefore, the point in which a circle with center (0, 4) and a radius of 3 intersects the line with equation y = 2x + 4 in the first quadrant is the point:
[tex]\displaystyle \Big(\frac{3\sqrt{5}}{5},\frac{6\sqrt5}{5}+4\Big)[/tex]
Or approximately:
[tex]\approx (1.342, 6.683)[/tex]
And we are finished!
