The line $y = x + 2$ intersects a circle with the equation $x^2 + y^2 = 4$. What type of solutions are possible and how many intersection points can there be?
Understand the Problem
The question describes a line intersecting a circle and asks about the nature and number of possible intersection points. We need to consider scenarios where the line intersects the circle at two points (two real solutions), is tangent to the circle (one real solution), or does not intersect the circle at all (no real solutions). The question tests your understanding of coordinate geometry and the relationship between linear and quadratic equations.
Answer
- Two intersections: $-\sqrt{2} < c < \sqrt{2}$ - One intersection: $c = \pm\sqrt{2}$ - No intersection: $c < -\sqrt{2}$ or $c > \sqrt{2}$
Answer for screen readers
- The line intersects the circle at two points if $-\sqrt{2} < c < \sqrt{2}$.
- The line is tangent to the circle at one point if $c = \pm\sqrt{2}$.
- The line does not intersect the circle if $c < -\sqrt{2}$ or $c > \sqrt{2}$.
Steps to Solve
- Express the line equation in terms of one variable
Given the line equation $y = x + c$, we want to substitute this into the equation of the circle to find the points of intersection.
- Substitute into the circle equation
Substitute $y = x + c$ into the circle equation $x^2 + y^2 = 1$: $x^2 + (x + c)^2 = 1$
- Expand and simplify the equation
Expanding and simplifying the equation gives: $x^2 + x^2 + 2cx + c^2 = 1$ $2x^2 + 2cx + c^2 - 1 = 0$
- Analyze the discriminant
The discriminant ($\Delta$) of the quadratic equation $ax^2 + bx + c = 0$ is given by $\Delta = b^2 - 4ac$. In our case, $a = 2$, $b = 2c$, and the constant term is $c^2 - 1$. So, the discriminant is: $\Delta = (2c)^2 - 4(2)(c^2 - 1)$ $\Delta = 4c^2 - 8(c^2 - 1)$ $\Delta = 4c^2 - 8c^2 + 8$ $\Delta = 8 - 4c^2$
- Determine conditions for different numbers of intersection points
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Two intersection points (two real solutions): $\Delta > 0$ $8 - 4c^2 > 0$ $4c^2 < 8$ $c^2 < 2$ $-\sqrt{2} < c < \sqrt{2}$
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One intersection point (one real solution, tangent): $\Delta = 0$ $8 - 4c^2 = 0$ $4c^2 = 8$ $c^2 = 2$ $c = \pm\sqrt{2}$
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No intersection points (no real solutions): $\Delta < 0$ $8 - 4c^2 < 0$ $4c^2 > 8$ $c^2 > 2$ $c < -\sqrt{2}$ or $c > \sqrt{2}$
- Match conditions to Options Listed
Now we match our findings with the provided options, but as the original options are not included, we just have the general solution for values of $c$.
- The line intersects the circle at two points if $-\sqrt{2} < c < \sqrt{2}$.
- The line is tangent to the circle at one point if $c = \pm\sqrt{2}$.
- The line does not intersect the circle if $c < -\sqrt{2}$ or $c > \sqrt{2}$.
More Information
The value of $c$ in $y = x + c$ determines the y-intercept of the line. As $c$ varies, the line shifts up and down. When $c$ is within the range $(-\sqrt{2}, \sqrt{2})$, the line intersects the unit circle at two points. When $c = \pm\sqrt{2}$, the line is tangent to the circle. Otherwise, the line does not intersect the circle.
Tips
- Algebraic errors: Mistakes in expanding $(x+c)^2$ or simplifying the discriminant can lead to an incorrect range of values for $c$.
- Incorrect interpretation of the discriminant: Misunderstanding the relationship between the discriminant's sign and the number of real roots.
- Sign errors: Watch out for sign errors when isolating $c$ after dealing with the discriminant inequality.
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