Equation of a circle described on the Latus rectum of the ellipse `x^2/25+y^2/16=1` as a diameter with centre on the +ve x-axis is

Options

A. `25x^2+25y^2-150 y -31=0`
B. `25x^2+25y^2-150x+31=0`
C. `25x^2+25y^2-150x-31=0`
D. `25x^2+25y^2-150y+31=0`

Answer

To find the equation of the circle described on the latus rectum of the ellipse \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \) as a diameter with its center on the positive x-axis, we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse is in the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where:
- \( a^2 = 25 \) (thus \( a = 5 \))
- \( b^2 = 16 \) (thus \( b = 4 \))

### Step 2: Calculate the eccentricity of the ellipse
The eccentricity \( e \) of the ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 3: Determine the foci of the ellipse
The foci of the ellipse are located at \( (\pm ae, 0) \):
\[ F_1 = (-3, 0) \quad \text{and} \quad F_2 = (3, 0) \]

### Step 4: Find the coordinates of the latus rectum
The latus rectum of the ellipse is vertical and passes through the foci. The coordinates of the endpoints of the latus rectum can be calculated using:
\[ \left( ae, \pm \frac{b^2}{a} \right) \]
Substituting the values:
\[ \left( 3, \pm \frac{16}{5} \right) \quad \text{and} \quad \left( -3, \pm \frac{16}{5} \right) \]
Thus, the points are:
\[ (3, \frac{16}{5}), (3, -\frac{16}{5}), (-3, \frac{16}{5}), (-3, -\frac{16}{5}) \]

### Step 5: Determine the diameter of the circle
The circle is described on the latus rectum as a diameter, so we will use the points \( (3, \frac{16}{5}) \) and \( (3, -\frac{16}{5}) \) as the endpoints of the diameter.

### Step 6: Find the center of the circle
The center of the circle is the midpoint of the diameter:
\[ \text{Center} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( 3, 0 \right) \]

### Step 7: Calculate the radius of the circle
The radius \( r \) is half the distance between the two points:
\[ r = \frac{1}{2} \times \text{distance between } (3, \frac{16}{5}) \text{ and } (3, -\frac{16}{5}) = \frac{1}{2} \times \left( \frac{16}{5} - (-\frac{16}{5}) \right) = \frac{1}{2} \times \frac{32}{5} = \frac{16}{5} \]

### Step 8: Write the equation of the circle
The equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 3 \), \( k = 0 \), and \( r = \frac{16}{5} \):
\[ (x - 3)^2 + (y - 0)^2 = \left(\frac{16}{5}\right)^2 \]
\[ (x - 3)^2 + y^2 = \frac{256}{25} \]

### Step 9: Multiply through by 25 to eliminate the fraction
\[ 25(x - 3)^2 + 25y^2 = 256 \]
Expanding this gives:
\[ 25(x^2 - 6x + 9) + 25y^2 = 256 \]
\[ 25x^2 - 150x + 225 + 25y^2 = 256 \]
\[ 25x^2 + 25y^2 - 150x - 31 = 0 \]

### Final Answer
The equation of the circle is:
\[ 25x^2 + 25y^2 - 150x - 31 = 0 \]