To determine the number of points of intersection between the parabola given by the equation y = x² and the circle centered at (0, 1) with a radius of 3, we’ll start by writing the equation of the circle. The standard form of a circle’s equation is given by:
(x - h)² + (y - k)² = r²
where (h, k) is the center of the circle and r is the radius. For our circle:
- Center:
(0, 1) - Radius:
3
Plugging in these values into the circle’s equation, we have:
(x - 0)² + (y - 1)² = 3²
which simplifies to:
x² + (y - 1)² = 9
Next, we substitute the equation of the parabola y = x² into the circle’s equation:
x² + (x² - 1)² = 9
Now, we will simplify and solve this equation step-by-step:
- Expand
(x² - 1)²: (x² - 1)(x² - 1) = x^4 - 2x² + 1- Substituting this back into the equation gives:
x² + x^4 - 2x² + 1 = 9- Simplifying further leads us to:
x^4 - x² + 1 - 9 = 0- which simplifies to:
x^4 - x² - 8 = 0
This is a quadratic in terms of u = x². Thus, we rewrite the equation:
u² - u - 8 = 0
To solve for u, we can use the quadratic formula:
u = rac{-b ± √(b² - 4ac)}{2a}
Where a = 1, b = -1, and c = -8. Plugging in these values:
u = rac{1 ± √(1 + 32)}{2} = rac{1 ± √33}{2}
Now, we have two potential values for u:
u₁ = rac{1 + √33}{2}u₂ = rac{1 - √33}{2}
The first value u₁ is positive, and since u = x², it will give us two corresponding values for x (one positive and one negative). The second value u₂ is negative, which does not yield any real solutions for x.
Thus, from u₁, representing u = x², we find:
x = ±√(u₁) = ±√ rac{1 + √33}{2}
Each of these two x-values will correspond to a unique y-value from the parabola. Therefore, we can conclude that the parabola and the circle intersect at:
2 points in total.
In summary, the parabola y = x² and the circle centered at (0, 1) with a radius of 3 intersect at two distinct points in the xy-plane.