+1713 I am doing "Equations" in Algebra 2 in Khanacdemy and stumbled on this problem:
Find all the Intersecting points of a circle and a line:
Circle equation :\((x-1)^2+(y-1)^2=1\)
Line equation :\(y=2x+1\)
And I tried the following but I am not sure if it is correct:
\(Since\ y=2x+1,\ we\ plug\ it\ into\ the\ circle\ equation. \\ (line\ 1): (x-1)^2+(y-1)^2=1 \\ (line\ 2): (x-1)^2+(2x+1-1)^2=1 \\ (line\ 3): (x-1)^2+(2x)^2=1 \\ (line\ 4): x^2-2x+1+4x^2=1 \\ (line\ 5): 5x^2-2x+1=1 \\ (line\ 6): 5x^2-2x=0 \\ (line\ 7): x(5x-2)=0 \\ x=0 \\ x=2/5\)
From there, I do not know what to do...
It would be great if you could help me from there and tell me if I am on the right path.
-Tommarvoloriddle
EDITS:
Problem Link:
https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:eq/x2ec2f6f830c9fb89:quad-sys/e/quadratic-systems?modal=1
Videos with the problem link:
Circle and line:
https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:eq/x2ec2f6f830c9fb89:quad-sys/v/systems-of-nonlinear-equations-3?modal=1
Parabola and line:
https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:eq/x2ec2f6f830c9fb89:quad-sys/v/line-and-parabola-system?modal=1
No solution:
https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:eq/x2ec2f6f830c9fb89:quad-sys/v/non-linear-systems-of-equations-3?modal=1
+2864 I am not sure on this one but I would just plug back in the original equations.
+26393 Find all the Intersecting points of a circle and a line:
Circle equation : (\(x-1)^2+(y-1)^2=1\)
Line equation : \(y=2x+1\)
\(\begin{array}{|rclrcl|} \hline \mathbf{x_1} &=& \mathbf{0} \\ && & y_1 &=& 2x_1+1 \\ && & y_1 &=& 2\cdot 0 +1 \\ && &\mathbf{ y_1} &=& \mathbf{ 1} \\\\ \mathbf{x_2} &=& \mathbf{0.4} \\ && & y_2 &=& 2x_2+1 \\ && & y_2 &=& 2\cdot 0.4 +1 \\ && &\mathbf{ y_2} &=& \mathbf{ 1.8 } \\ \hline \end{array} \)
The two Intersecting points of a circle and a line are: \(\mathbf{(0,\ 1)}\) and \(\mathbf{(0.4,\ 1.8)}\)
