This solution will guide you through the process of solving a classic coordinate geometry problem involving ellipses, tangents, and circles. We'll break down the problem into manageable steps, explaining the 'why' behind each action, and leveraging key mathematical concepts.

Key Concepts and Strategic Approach

To solve this problem effectively, we will utilize the following fundamental principles:

1. Standard Equation of an Ellipse: An ellipse centered at the origin is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Here, $a$ is the semi-major axis length and $b$ is the semi-minor axis length.
2. Parametric Form of a Tangent to an Ellipse: The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at a general point $P(a \cos\theta, b \sin\theta)$ is $\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$. This form is highly advantageous for representing any tangent to the ellipse without specifying a particular point.
3. Tangents at Major Axis Extremities: For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the major axis lies along the x-axis (if $a>b$). Its endpoints are $(\pm a, 0)$, and the tangents at these points are the vertical lines $x = a$ and $x = -a$.
4. Equation of a Circle with Diameter Endpoints: If $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of a diameter of a circle, its equation is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. This form directly uses the property that the angle subtended by a diameter at any point on the circumference is $90^\circ$.
5. Trigonometric Identities: Specifically, half-angle formulas and reciprocal identities are crucial for simplifying expressions:
   • $1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)$
   • $1 - \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)$
   • $\sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$
   • $\tan\left(\frac{\theta}{2}\right) \cdot \cot\left(\frac{\theta}{2}\right) = 1$
   • $\tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A}$

Our strategy will be:

1. Convert the given ellipse equation to its standard form to identify its semi-axes.
2. Write the general equation of a tangent to this ellipse using parametric coordinates.
3. Determine the equations of the vertical tangents at the major axis extremities.
4. Find the intersection points (B and C) of the general tangent with these vertical tangents.
5. Simplify the coordinates of B and C using trigonometric identities to make subsequent calculations easier.
6. Formulate the equation of the circle that has BC as its diameter.
7. Identify the fixed point(s) through which this circle passes, independent of the parameter $\theta$.

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Step 1: Standardizing the Ellipse Equation and Identifying Parameters

The given equation of the ellipse is $x^2 + 4y^2 = 4$. Why this step? To apply standard formulas for tangents and to understand the geometry of the ellipse (like its major axis and vertices), we must first convert its equation into the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Divide the entire equation by 4: $\frac{x^2}{4} + \frac{4y^2}{4} = \frac{4}{4}$ $\frac{x^2}{4} + \frac{y^2}{1} = 1$ Comparing this with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: We find $a^2 = 4 \implies a = 2$ (length of the semi-major axis). And $b^2 = 1 \implies b = 1$ (length of the semi-minor axis). Since $a > b$, the major axis lies along the x-axis.

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Step 2: Finding the Equation of the Tangent to the Ellipse at a General Point

Why this step? The problem involves "a tangent to the ellipse". To represent this general tangent, the parametric form is most convenient because it covers all possible tangents by varying the parameter $\theta$.

A general point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $(a \cos\theta, b \sin\theta)$. For our ellipse, $a=2$ and $b=1$, so a general point of tangency $P$ is $(2\cos\theta, \sin\theta)$.

The equation of the tangent at this point is: $\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$ Substituting $a=2$ and $b=1$: $\frac{x}{2}\cos\theta + \frac{y}{1}\sin\theta = 1$ To clear the denominator, multiply the entire equation by 2: $x\cos\theta + 2y\sin\theta = 2 \quad (*)$ This is the equation of the general tangent.

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Step 3: Determining the Tangents at the Extremities of the Major Axis

Why this step? The problem states that the general tangent meets "the tangents at the extremities of its major axis". We need the equations of these specific tangents to find the intersection points.

For our ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$, the semi-major axis $a=2$. The major axis lies along the x-axis, so its extremities (vertices) are $(\pm a, 0)$, which are $(2, 0)$ and $(-2, 0)$.

The tangents at these points are vertical lines:

1. Tangent at $(2, 0)$ is $x = 2$.
2. Tangent at $(-2, 0)$ is $x = -2$.

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Step 4: Finding the Coordinates of Points B and C

Why this step? Points B and C are defined as the intersection points of the general tangent with the major axis extremity tangents. These points will be the endpoints of the diameter of the circle we need to analyze.

To find point B: Point B is the intersection of the general tangent ($x\cos\theta + 2y\sin\theta = 2$) and the line $x = -2$. Substitute $x = -2$ into the tangent equation: $(-2)\cos\theta + 2y_B\sin\theta = 2$ Solve for $y_B$: $2y_B\sin\theta = 2 + 2\cos\theta$ $y_B = \frac{2(1 + \cos\theta)}{2\sin\theta}$ $y_B = \frac{1 + \cos\theta}{\sin\theta}$ So, $B = \left(-2, \frac{1 + \cos\theta}{\sin\theta}\right)$.

To find point C: Point C is the intersection of the general tangent ($x\cos\theta + 2y\sin\theta = 2$) and the line $x = 2$. Substitute $x = 2$ into the tangent equation: $(2)\cos\theta + 2y_C\sin\theta = 2$ Solve for $y_C$: $2y_C\sin\theta = 2 - 2\cos\theta$

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