Answer:
The point is
![((3)/(5)√(5),(6)/(5)√(5))](https://img.qammunity.org/2020/formulas/mathematics/high-school/bhgq1sx8ripcw5kfefvfldichpn96ibz5u.png)
Explanation:
∵ The circle has center (0 , 0) and radius 3
∵ The equation of any circle is
![(x-h)^(2)+(y-k)^(2)=r^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pef6vup9m8fe069mn7ia9hgp8m0ly3p5um.png)
where (h , k) is the center of the circle and r is the radius of it
∴
![x^(2)+y^(2)=3^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/sldrr8dz48n5qrcveewccis8urkvh2c6ye.png)
∴
![x^(2)+y^(2)=9](https://img.qammunity.org/2020/formulas/mathematics/high-school/5ggv6qfu739vwua16gnnv7xulqzkzoa45i.png)
∵ The equation of the line is y = 2x
We will substitute y in the equation of the circle by 2x
∴
![x^(2)+(2x)^(2)=9](https://img.qammunity.org/2020/formulas/mathematics/high-school/7vu629r8i36dwzwp1wpby2wb7e2v3y98m0.png)
∴
![x^(2)+4x^(2)=9](https://img.qammunity.org/2020/formulas/mathematics/high-school/endo2pwlphqzgj64jnelct7874mdpq71ve.png)
∴
![x^(2)=(9)/(5)](https://img.qammunity.org/2020/formulas/mathematics/high-school/fyl9w3tedlifnrpw0dmteysy0s89zvf7rj.png)
∴
![x=\sqrt{(9)/(5)}=(3)/(√(5))((√(5))/(√(5)))=(3)/(5)√(5)](https://img.qammunity.org/2020/formulas/mathematics/high-school/tuccfkdm9qtcrh3vfckzt0brru2fdjcd45.png)
∴ y =
![2((3)/(5)√(5))=(6)/(5)√(5)](https://img.qammunity.org/2020/formulas/mathematics/high-school/9o3mcu2tp6aku5aznxwy6wtm159pq2n12m.png)
The line intersects the circle at point
![((3)/(5)√(5),(6)/(5)√(5))](https://img.qammunity.org/2020/formulas/mathematics/high-school/bhgq1sx8ripcw5kfefvfldichpn96ibz5u.png)
The decimal values (1.342 , 2.684)