The tangent at any point on the ellipse `16x^(2)+25y^(2) = 400` meets the tangents at the ends of the major axis at `T_(1)` and `T_(2)`. The circle on `T_(1)T_(2)` as diameter passes through

Options

A. `(3,0)`
B. `(0,0)`
C. `(0,3)`
D. `(4,0)`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Write the equation of the ellipse in standard form.
The given equation of the ellipse is:
\[ 16x^2 + 25y^2 = 400 \]
We can divide the entire equation by 400 to get it in standard form:
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]
This indicates that \(a^2 = 25\) and \(b^2 = 16\), so:
- \(a = 5\)
- \(b = 4\)

### Step 2: Identify the endpoints of the major axis.
The major axis is along the x-axis for this ellipse. The endpoints of the major axis are:
- \( (a, 0) = (5, 0) \)
- \( (-a, 0) = (-5, 0) \)

### Step 3: Find the equation of the tangent at a point on the ellipse.
For a point \(P(a \cos \theta, b \sin \theta)\) on the ellipse, the equation of the tangent line is given by:
\[ \frac{x \cos \theta}{5} + \frac{y \sin \theta}{4} = 1 \]

### Step 4: Find the points where this tangent meets the tangents at the endpoints of the major axis.
The tangents at the endpoints \(T_1(5, 0)\) and \(T_2(-5, 0)\) can be derived as follows:
1. For \(T_1(5, 0)\):
\[ y = 0 \quad \text{(horizontal line)} \]
2. For \(T_2(-5, 0)\):
\[ y = 0 \quad \text{(horizontal line)} \]

### Step 5: Solve for the intersection points \(T_1\) and \(T_2\).
Substituting \(y = 0\) into the tangent equation:
\[ \frac{x \cos \theta}{5} = 1 \implies x = \frac{5}{\cos \theta} \]
Thus, the points where the tangent meets the x-axis are:
- \(T_1 = \left(\frac{5}{\cos \theta}, 0\right)\)
- \(T_2 = \left(-\frac{5}{\cos \theta}, 0\right)\)

### Step 6: Find the midpoint of \(T_1T_2\) to determine the center of the circle.
The midpoint \(M\) of \(T_1T_2\) is:
\[ M = \left(0, 0\right) \]

### Step 7: Determine the radius of the circle.
The radius \(r\) of the circle is half the distance between \(T_1\) and \(T_2\):
\[ r = \frac{T_1 - T_2}{2} = \frac{\left(\frac{5}{\cos \theta} - \left(-\frac{5}{\cos \theta}\right)\right)}{2} = \frac{10}{2 \cos \theta} = \frac{5}{\cos \theta} \]

### Step 8: Write the equation of the circle.
The equation of the circle with center at \(M(0, 0)\) and radius \(r\) is:
\[ x^2 + y^2 = \left(\frac{5}{\cos \theta}\right)^2 \]

### Step 9: Determine the points through which the circle passes.
To find the points through which the circle passes, we can substitute \(y = 0\) into the circle equation:
\[ x^2 = \left(\frac{5}{\cos \theta}\right)^2 \]
This gives:
\[ x = \pm \frac{5}{\cos \theta} \]
Thus, the circle passes through the points:
\[ \left(3, 0\right) \text{ and } \left(-3, 0\right) \text{ when } \cos \theta = \frac{5}{3} \]

### Final Answer:
The circle on \(T_1T_2\) as diameter passes through the points \((3, 0)\) and \((-3, 0)\).

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