The equation of the circle passing through the foci of the ellipse x²/16 + y²/9 = 1, and having cent

Q: Wait, why do we use $(\pm ae, 0)$ for the foci? Shouldn't it be $(\pm a/e, 0)$?

That is a common confusion! Remember that $e < 1$ for an ellipse. If you used $a/e$, you would be placing the foci outside the ellipse, which is impossible. The foci are always inside the ellipse, so we multiply by $e$ to bring them closer to the center. Always keep the inequality $e < 1$ in mind as a sanity check.

Q: Do I need to calculate the distance for both foci to find the radius?

Great question! Because the ellipse is symmetric about both axes and the circle's center $(0, 3)$ lies on the y-axis (the minor axis of the ellipse), the distance to both foci will be identical. You only need to calculate it for one focus. Symmetry is your best friend in JEE geometry—use it to save time!

Q: Why can't I just plug the foci into the options and see which one works?

You absolutely can! In a high-pressure exam like JEE Advanced, if you are stuck, checking the options is a valid strategy. However, understanding the derivation is crucial for building the intuition needed for harder problems where options might not be so easily verifiable. Use the derivation to learn, and the options to verify.

Question

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

Options:

(A) $x^2 + y^2 - 6y - 7 = 0$

(B) $x^2 + y^2 - 6y + 7 = 0$

(C) $x^2 + y^2 - 6y - 5 = 0$

(D) $x^2 + y^2 - 6y + 5 = 0$

Sigma Percentile · Accepted Answer

### The Dance of Conics: An Ellipse and a Circle

Welcome, future engineer. Today, we aren't just solving a coordinate geometry problem; we are choreographing a dance between two of the most elegant shapes in mathematics: the ellipse and the circle.

When you look at an equation like

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

I don't want you to see just numbers and variables. I want you to see a physical object—a stretched circle, an oval of perfect proportions. Let's peel back the layers of this problem together.

### Phase 1: Decoding the Ellipse

Every journey begins with understanding our terrain. We are given the ellipse

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

By comparing this to the standard form

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

we immediately identify our parameters. Here, $a^2 = 16$, which gives us a semi-major axis $a = 4$. Similarly, $b^2 = 9$, giving us a semi-minor axis $b = 3$.

But why do we care about $a$ and $b$? Because they dictate the 'stretch' of the ellipse, which is quantified by the eccentricity, $e$.

The eccentricity is the soul of the ellipse; it tells us how far the foci are from the center. We use the classic relation:

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

Substituting our values, we get:

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

### Phase 2: Finding the Anchor Points

Now, we need to locate the foci. These are the two points that define the ellipse's very existence.

For an ellipse centered at the origin, the foci lie on the major axis at $(\pm ae, 0)$. Let's calculate this: $a \cdot e = 4 \cdot \frac{\sqrt{7}}{4}$.

Look at that! The fours cancel out with such satisfying precision, leaving us with $\pm \sqrt{7}$. So, our foci are located at $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$.

These are the points our circle must pass through. Imagine these as two nails in a board, and we are drawing a circle that must hit both of them.

### Phase 3: The Circle's Geometry

Now, shift your perspective. We are no longer thinking about the ellipse; we are thinking about a circle. We are told the center of this circle is at $(0, 3)$.

The standard equation of a circle with center $(h, k)$ and radius $R$ is $(x - h)^2 + (y - k)^2 = R^2$. Substituting our center $(0, 3)$, we get:

$$x^2 + (y - 3)^2 = R^2$$

We are missing one crucial piece: the radius $R$. But we have a clue! The circle passes through the foci.

This means the distance from the center $(0, 3)$ to the focus $(\sqrt{7}, 0)$ must be the radius $R$. Let's use the distance formula:

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

Calculating this, we get $R^2 = 7 + 9 = 16$. The radius is $4$. It is beautiful how the numbers align, isn't it?

### Phase 4: The Final Synthesis

We are at the finish line. We have the equation $x^2 + (y - 3)^2 = 16$. To match the format of our options, we need to expand this.

Let's expand the term $(y - 3)^2$ using the identity $(a - b)^2 = a^2 - 2ab + b^2$. This gives us:

$$x^2 + (y^2 - 6y + 9) = 16$$

Finally, we bring the $16$ to the left side: $x^2 + y^2 - 6y + 9 - 16 = 0$. Simplifying the constants, $9 - 16$ gives us $-7$.

Thus, our final equation is:

$$\mathbf{x^2 + y^2 - 6y - 7 = 0}$$

Take a moment to appreciate this. We started with an abstract ellipse, found its hidden foci, used those to define a circle, and arrived at a clean, elegant algebraic expression. This is the power of coordinate geometry—it turns visual intuition into rigorous proof.

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

Select Answer:

Visualized Solution

Analyze the Ellipse Equation

Extracting $a^{2}$ and $b^{2}$

Formula for Eccentricity $e$

Calculating Eccentricity

Coordinates of the Foci

The Circle's Center

Standard Equation of Circle

Radius from Center to Focus

Calculating $R^{2}$

Visualizing the Circle

Expanding the Circle Equation

Final Circle Equation

The Sigma Insight: Foci, Directrices, and Eccentricity

The Dance of Conics

Phase 1

Phase 2

Phase 3

Phase 4

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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 equation of the circle passing through the foci of the ellipse 16x2​+9y2​=1, and having centre at (0,3) is