Answer:
The point is [tex](\frac{3}{5}\sqrt{5},\frac{6}{5}\sqrt{5})[/tex]
Step-by-step explanation:
∵ The circle has center (0 , 0) and radius 3
∵ The equation of any circle is [tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]
where (h , k) is the center of the circle and r is the radius of it
∴ [tex]x^{2}+y^{2}=3^{2}[/tex]
∴ [tex]x^{2}+y^{2}=9[/tex]
∵ The equation of the line is y = 2x
We will substitute y in the equation of the circle by 2x
∴ [tex]x^{2}+(2x)^{2}=9[/tex]
∴ [tex]x^{2}+4x^{2}=9[/tex]
∴ [tex]x^{2}=\frac{9}{5}[/tex]
∴[tex]x=\sqrt{\frac{9}{5}}=\frac{3}{\sqrt{5}}(\frac{\sqrt{5}}{\sqrt{5}})=\frac{3}{5}\sqrt{5}[/tex]
∴ y = [tex]2(\frac{3}{5}\sqrt{5})=\frac{6}{5}\sqrt{5}[/tex]
The line intersects the circle at point [tex](\frac{3}{5}\sqrt{5},\frac{6}{5}\sqrt{5})[/tex]
The decimal values (1.342 , 2.684)
