Find the radius of the circle passing through the foci of an ellipse `9x^2+16y^2=144` and having least radius.

Answer

To find the radius of the circle passing through the foci of the ellipse given by the equation \(9x^2 + 16y^2 = 144\) and having the least radius, we will follow these steps:

### Step 1: Convert the equation of the ellipse to standard form
We start with the given equation of the ellipse:
\[ 9x^2 + 16y^2 = 144 \]
To convert this to standard form, we divide every term by 144:
\[ \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \]
This simplifies to:
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
Now, we can identify \(a^2\) and \(b^2\):
\[ a^2 = 16 \quad \text{and} \quad b^2 = 9 \]
Thus, we have:
\[ a = 4 \quad \text{and} \quad b = 3 \]

### Step 2: Calculate the eccentricity (e) of the ellipse
The eccentricity \(e\) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ e = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{16 - 9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

### Step 3: Find the coordinates of the foci
The foci of the ellipse are located at \((\pm ae, 0)\). Thus, we calculate:
\[ ae = 4 \cdot \frac{\sqrt{7}}{4} = \sqrt{7} \]
Therefore, the coordinates of the foci are:
\[ (\sqrt{7}, 0) \quad \text{and} \quad (-\sqrt{7}, 0) \]

### Step 4: Determine the radius of the circle
The radius of the circle that passes through the foci and has the least radius is equal to the distance from the center of the ellipse (which is at the origin \((0, 0)\)) to either of the foci. The distance from the center to one focus is given by:
\[ \text{Distance} = \sqrt{(\sqrt{7} - 0)^2 + (0 - 0)^2} = \sqrt{7} \]

### Final Answer
Thus, the radius of the circle passing through the foci of the ellipse and having the least radius is:
\[ \text{Radius} = \sqrt{7} \]

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