Question

Choose ,begin emphasis,all,end emphasis, of the points where the graphs of the line y equals x minus 3 and the circle left-parenthesis x minus 1 right-parenthesis squared plus y squared equals 4 intersect.

Answer options with 5 options
A.
the point 0 comma negative 3

B.
the point 0 comma 3

C.
the point 1 comma negative 2

D.
the point 1 comma 0

E.
the point 3 comma 0

Answer 1

To find the points of intersection between the line \( y = x - 3 \) and the circle \( (x - 1)^2 + y^2 = 4 \), we can substitute the expression for \( y \) from the line equation into the circle equation.

1. The line equation is:
\[
y = x - 3
\]

2. Substitute \( y \) into the circle equation:
\[
(x - 1)^2 + (x - 3)^2 = 4
\]

3. Expand and simplify:
\[
(x - 1)^2 = x^2 - 2x + 1
\]
\[
(x - 3)^2 = x^2 - 6x + 9
\]

Thus,
\[
(x - 1)^2 + (x - 3)^2 = (x^2 - 2x + 1) + (x^2 - 6x + 9) = 2x^2 - 8x + 10
\]

4. Set the equation equal to 4:
\[
2x^2 - 8x + 10 = 4
\]

5. Simplify:
\[
2x^2 - 8x + 6 = 0
\]
\[
x^2 - 4x + 3 = 0
\]

6. Factor the quadratic:
\[
(x - 3)(x - 1) = 0
\]

7. Solve for \( x \):
\[
x = 3 \quad \text{or} \quad x = 1
\]

8. Find corresponding \( y \) values:
- For \( x = 3 \):
\[
y = 3 - 3 = 0 \quad \Rightarrow \quad (3, 0)
\]
- For \( x = 1 \):
\[
y = 1 - 3 = -2 \quad \Rightarrow \quad (1, -2)
\]

The points of intersection are \( (3, 0) \) and \( (1, -2) \).

Thus, the points where the graphs intersect are:

- **C**. the point \( 1, -2 \)
- **E**. the point \( 3, 0 \)

**Final answer: (C and E)**.