Question

Choose all of the points where the graphs of the line y = x - 3 and the circle (x - 1) ^ 2 + y ^ 2 = 4 intersect

Answer 1

To solve the exercise, first replace the value of y of the line in the equation of the circle, like this

`\begin{gathered} y=x-3 \\ (x-1)^2+y^2=4 \end{gathered}`
`(x-1)^2+(x-3)^2=4`

Solving for x you have

`\begin{gathered} (x-1)(x-1)+(x-3)(x-3)=4 \\ x^2-x-x+1+x^2-3x-3x+9=4 \end{gathered}`

Subtract 4 from both of the equation and operate like terms

`\begin{gathered} x^2-x-x+1+x^2-3x-3x+9-4=4-4 \\ 2x^2-8x+6=0 \end{gathered}`

Now you can use the general formula for quadratic equations, that is

`\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}`

In this case

`\begin{gathered} 2x^2-8x+6=0 \\ a=2 \\ b=-8 \\ c=6 \end{gathered}`

So, you have

`\begin{gathered} x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(2)(6)}}{2(2)} \\ x=\frac{8\pm\sqrt[]{64-48}}{4} \\ x=\frac{8\pm\sqrt[]{16}}{4} \\ x=(8\pm4)/(4) \end{gathered}`

Then the solutions of the quadratic equation will be

`\begin{gathered} x_1=(8+4)/(4) \\ x_1=(12)/(4) \\ x_1=3 \end{gathered}`
`\begin{gathered} x_2=(8-4)/(4) \\ x_2=(4)/(4) \\ x_2=1 \end{gathered}`

Finally, to get the y-coordinate of the intersection points between the line and the circle, replace the values of x found in any of the initial equations, for example in the equation of the line

`\begin{gathered} y_1=x_1-3 \\ y_1=3-3 \\ y_1=0 \end{gathered}`
`\begin{gathered} y_2=x_2-3 \\ y_2=1-3 \\ y_2=-2 \end{gathered}`

Therefore, the points where the line and the circle intersect are

`(3,0)\text{ and }(1,-2)`

And the correct answers are E. and C.