nearest distance of y=\left(x-5\right)^{2\ }+1 to 4=\ x^{2\ }+\ y^{2}

We are tasked with finding the nearest distance between the parabola \( y = (x - 5)^2 + 1 \) and the circle \( x^2 + y^2 = 4 \).

### Step 1: Parametrize the Circle
The equation of the circle is \( x^2 + y^2 = 4 \), which is a circle centered at the origin with radius 2. We can parametrize the circle using:
\[
x = 2 \cos \theta, \quad y = 2 \sin \theta
\]
where \( \theta \) is the parameter that varies between \( 0 \) and \( 2\pi \).

### Step 2: Expression for the Distance
The distance \( d \) between a point on the circle and the parabola can be written as:
\[
d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}
\]
where \( (x_1, y_1) \) represents a point on the circle and \( (x_2, y_2) \) represents a point on the parabola.

Thus, given the parametric equations of the circle, the goal is to minimize the distance between a point \( (x_1, y_1) = (2\cos\theta, 2\sin\theta) \) on the circle and the point on the parabola. Therefore, our task is to minimize this function of \( \theta \).

Would you like me to continue by deriving the distance formula explicitly and finding the critical points?

nearest distance of y=(x-5)^2 + 1 to 4 = x^2 + y^2

This solution explains how to find the nearest distance between the parabola y = (x - 5)^2 + 1 and the circle x^2 + y^2 = 4. Learn about parametric equations, the distance formula, and techniques for minimizing distances between geometric curves. Suitable for high school students in Grades 11-12.

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