Given a unit circle centered at the origin, find the points where the circle intersects the line $y = 2x + 3$.
Answer 1
Amelia Mitchell
To find the points of intersection between the line $y = 2x + 3$ and the unit circle $x^2 + y^2 = 1$, we substitute $y = 2x + 3$ into the circle’s equation:
$ x^2 + (2x + 3)^2 = 1 $
Expanding and simplifying:
$ x^2 + (4x^2 + 12x + 9) = 1 $
$ 5x^2 + 12x + 9 = 1 $
$ 5x^2 + 12x + 8 = 0 $
Using the quadratic formula where $a = 5, b = 12, c = 8$:
$ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} $
$ x = \frac{-12 \pm \sqrt{144 – 160}}{10} $
$ x = \frac{-12 \pm \sqrt{-16}}{10} $
Since the discriminant is negative, there are no real solutions, meaning the unit circle and the line $y = 2x + 3$ do not intersect.
Answer 2
Maria Rodriguez
To determine the intersection points between the circle $x^2 + y^2 = 1$ and the line $y = 2x + 3$, we start by substituting $2x + 3$ for $y$ in the circle’s equation:
$ x^2 + (2x + 3)^2 = 1 $
Expanding and collecting like terms:
$ x^2 + 4x^2 + 12x + 9 = 1 $
$ 5x^2 + 12x + 9 = 1 $
$ 5x^2 + 12x + 8 = 0 $
Applying the quadratic formula $x = frac{-b pm sqrt{b^2 – 4ac}}{2a}$ where $a = 5, b = 12, c = 8$:
$ x = frac{-12 pm sqrt{144 – 160}}{10} $
$ x = frac{-12 pm sqrt{-16}}{10} $
Since the discriminant $(-16)$ is negative, there are no intersections (no real solutions).
Answer 3
Benjamin Clark
We substitute $y = 2x + 3$ into $x^2 + y^2 = 1$:
$ x^2 + (2x + 3)^2 = 1 $
$ x^2 + 4x^2 + 12x + 9 = 1 $
$ 5x^2 + 12x + 8 = 0 $
Using the quadratic formula:
$ x = frac{-12 pm sqrt{-16}}{10} $
No real solutions since the discriminant is negative.
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