The radius of the circle passing through the foci of the ellipse x²/16 + y²/9 = 1, and having its ce

Q: Wait, why is the ellipse horizontal? How do I know if it's vertical?

Great question! In the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, if $a^2 > b^2$, the ellipse is stretched along the x-axis, making it horizontal. If $b^2 > a^2$, it would be stretched along the y-axis, making it vertical. Here, $16 > 9$, so it is definitely horizontal!

Q: Can I use the other focus $(-\sqrt{7}, 0)$ instead of $(\sqrt{7}, 0)$?

Absolutely! Because the center of the circle is at $(0, 3)$ and the ellipse is symmetric about the y-axis, the distance from $(0, 3)$ to $(\sqrt{7}, 0)$ is identical to the distance from $(0, 3)$ to $(-\sqrt{7}, 0)$. You will get the same radius either way.

Q: Why can't we just use the distance from the center to the vertex?

The problem specifically states the circle passes through the *foci*, not the vertices. The vertices are at $(\pm 4, 0)$, but the foci are at $(\pm \sqrt{7}, 0)$. Using the wrong points will lead to an incorrect radius. Always stick to the points defined in the problem statement!

Question

The radius of the circle passing through the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$, and having its centre at $(0, 3)$ is

Options:

(A) 4

(B) 3

(C) $\sqrt{12}$

(D) $7/2$

Sigma Percentile · Accepted Answer

### Analyzing the Ellipse Geometry

Imagine you are standing in a vast, silent observatory, looking at the celestial mechanics of an ellipse. The equation

$$\frac{x^2}{16} + \frac{y^2}{9} = 1$$

is not just a collection of numbers; it is a blueprint of a perfect, flattened circle. In the world of JEE Advanced, visualizing this geometry is your first step toward mastery.

We see a horizontal ellipse, gracefully centered at the origin $(0,0)$. By comparing this to the standard form

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

we immediately identify our semi-axes: $a^2 = 16$, which gives us $a = 4$, and $b^2 = 9$, which gives us $b = 3$. This is the foundation of our journey.

### The Hunt for the Foci

Now, we must find the heart of the ellipse—the foci. The eccentricity, $e$, is the measure of how 'flat' our ellipse is. It is the soul of the conic section.

We use the elegant formula:

$$e = \sqrt{1 - \frac{b^2}{a^2}}$$

Substituting our values, we get:

$$e = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{16 - 9}{16}} = \frac{\sqrt{7}}{4}$$

This value tells us exactly how far the foci are from the center. For a horizontal ellipse, the foci reside at $(\pm ae, 0)$. Multiplying $a = 4$ by $e = \frac{\sqrt{7}}{4}$, the fours cancel out with poetic precision, leaving us with the coordinates $F_1(\sqrt{7}, 0)$ and $F_2(-\sqrt{7}, 0)$.

### The Geometric Bridge

We are now given a circle with its center at $C(0, 3)$. The problem states this circle passes through the foci $F_1(\sqrt{7}, 0)$ and $F_2(-\sqrt{7}, 0)$.

The radius $R$ is the constant distance from the center $C$ to any point on the circle's edge. Since the circle touches the focus $F_1(\sqrt{7}, 0)$, the radius is simply the distance between $C(0, 3)$ and $F_1(\sqrt{7}, 0)$.

We reach for our trusty distance formula:

$$R = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

### The Final Revelation

Let us perform the final calculation with care. Substituting the coordinates, we have:

$$R = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2}$$

This simplifies to:

$$R = \sqrt{(\sqrt{7})^2 + (-3)^2} = \sqrt{7 + 9} = \sqrt{16}$$

The result is a clean, satisfying **$R = 4$**. Throughout this problem, we have moved from the abstract equation of an ellipse to the concrete reality of a circle's radius. It is a beautiful reminder that in mathematics, every step is a logical bridge connecting one truth to the next.

Animated Solution for Mathematics - Conic Sections: The radius of the circle passing through the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ , and having its centre at $(0, 3)$ is

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Visualized Solution

Visualize the Ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$

Identify Semi-axes $a$ and $b$

Formula for Eccentricity $e$

Calculate Eccentricity $e$

Locate the Foci $(\pm a e, 0)$

Calculate Foci Coordinates

Define the Circle at $C (0, 3)$

The Distance Formula for Radius $R$

Substitute Coordinates into Formula

Final Calculation of $R$

Summary and Final Answer

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Ellipse Geometry

The Hunt for the Foci

The Geometric Bridge

The Final Revelation

Similar Questions

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The radius of the circle passing through the foci of the ellipse 16x2​+9y2​=1, and having its centre at (0,3) is

Select Answer:

Visualized Solution

Visualize the Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 16x2​+9y2​=1

Identify Semi-axes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Formula for Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e

Calculate Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e

Locate the Foci .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (±ae,0)