Key Concepts and Formulas

Before we dive into the solution, let's recall the fundamental concepts and formulas we'll be using. A strong understanding of these will make solving problems like this much easier.

1. Standard Equation of an Ellipse: For an ellipse centered at the origin, its standard equation is given by: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

   • Here, $a$ and $b$ represent the lengths of the semi-axes.
   • Orientation: To determine the major and minor axes:
     • If $a^2 > b^2$ (meaning $a > b$), then $a$ is the length of the semi-major axis, and the major axis lies along the x-axis.
     • If $b^2 > a^2$ (meaning $b > a$), then $b$ is the length of the semi-major axis, and the major axis lies along the y-axis.
   • The center of this ellipse is at $(0,0)$.

2. Eccentricity ($e$) of an Ellipse: The eccentricity quantifies how "stretched out" or "flat" an ellipse is. It's a key parameter for locating the foci.

   • It's always defined such that $0 < e < 1$.
   • The general formula is $e = \sqrt{1 - \frac{(\text{semi-minor axis})^2}{(\text{semi-major axis})^2}}$.
   • Specifically:
     • If the major axis is along the x-axis ($a>b$), then $e = \sqrt{1 - \frac{b^2}{a^2}}$.
     • If the major axis is along the y-axis ($b>a$), then $e = \sqrt{1 - \frac{a^2}{b^2}}$.

3. Foci of an Ellipse: The foci are two fixed points on the major axis of the ellipse, fundamental to its definition. Their coordinates depend on the orientation of the major axis:

   • If the major axis is along the x-axis (i.e., $a>b$), the foci are at $(\pm ae, 0)$.
   • If the major axis is along the y-axis (i.e., $b>a$), the foci are at $(0, \pm be)$ where $b$ is the semi-major axis, or equivalently, $(0, \pm ae)$ where $a$ is the semi-major axis as per the definition in point 1. To avoid confusion, it's best to remember the foci are at $(\pm \text{distance from center}, 0)$ or $(0, \pm \text{distance from center})$ where the distance is $e \times (\text{semi-major axis})$. For our problem, we will use $ae$ as the distance from the center along the major axis.

4. Standard Equation of a Circle: A circle with center $(h,k)$ and radius $r$ has the equation: $(x-h)^2 + (y-k)^2 = r^2$

5. Distance Formula: The distance $d$ between any two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

---

Step-by-Step Solution

Our objective is to find the equation of a circle. To achieve this, we need two pieces of information: its center and its radius. The problem provides the center directly, but we need to calculate the radius using the fact that the circle passes through the foci of the given ellipse.

Step 1: Analyze the Given Ellipse and Find its Foci

The equation of the ellipse is given as: $\frac{x^2}{16} + \frac{y^2}{9} = 1$

• Identify $a^2$ and $b^2$:

  • Why this step? The first step in working with an ellipse is to identify its fundamental dimensions. By comparing the given equation with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we can directly identify the values of the squares of the semi-axes, $a^2$ and $b^2$.
  • From the given equation, we have $a^2 = 16$ and $b^2 = 9$.

• Determine the Orientation of the Major Axis:

  • Why this step? The orientation (whether the major axis is along the x-axis or y-axis) dictates which formulas to use for calculating eccentricity and the coordinates of the foci. It's crucial for correctly locating the foci.
  • Since $16 > 9$, we have $a^2 > b^2$. This means that $a$ is the semi-major axis, and the major axis lies along the x-axis.

• Calculate $a$ and $b$:

  • Why this step? We need the actual lengths of the semi-axes ($a$ and $b$) to calculate the eccentricity and the precise coordinates of the foci.
  • $a = \sqrt{16} = 4$
  • $b = \sqrt{9} = 3$

• Calculate the Eccentricity ($e$):

  • Why this step? The eccentricity is a crucial intermediate value. It tells us how far the foci are from the center of the ellipse. We need $e$ to find the exact coordinates of the foci.
  • Since the major axis is along the x-axis, we use the formula: $e = \sqrt{1 - \frac{b^2}{a^2}}$
  • Substitute the values of $a^2$ and $b^2$: $e = \sqrt{1 - \frac{9}{16}}$ $e = \sqrt{\frac{16-9}{16}}$ $e = \sqrt{\frac{7}{16}}$ $e = \frac{\sqrt{7}}{4}$

• Determine the Foci Coordinates:

  • Why this step? The problem states that the circle passes through these foci. Therefore, finding their exact coordinates is essential because these points will lie on the circumference of the circle, allowing us to determine the circle's radius.
  • For an ellipse with the major axis along the x-axis and centered at the origin, the foci are located at $(\pm ae, 0)$.
  • Calculate the product $ae$: $ae = 4 \times \frac{\sqrt{7}}{4} = \sqrt{7}$
  • Therefore, the coordinates of the foci are $F_1 = (\sqrt{7}, 0)$ and $F_2 = (-\sqrt{7}, 0)$.

Step 2: Determine the Radius of the Circle

We are given that the center of the circle is $C = (0,3)$. We also know that the circle passes through the foci, which we found to be $F_1 = (\sqrt{7}, 0)$ and $F_2 = (-\sqrt{7}, 0)$.

• Why this step? To write the equation of a circle, we need its radius. The radius is defined as the distance from the center of the circle to any point on its circumference. Since the foci lie on the circumference of the circle, we can calculate the distance between the circle's center and one of the foci to find the radius.
• Apply the Distance Formula: Let's find the distance $r$ between the center $C(0,3)$ and one of the foci, say $F_1(\sqrt{7}, 0)$. $r = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2}$
  • Explanation: We are using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ with $(x_1, y_1) = (0,3)$ and $(x_2, y_2) = (\sqrt{7}, 0)$. $r = \sqrt{(\sqrt{7})^2 + (-3)^2}$ $r = \sqrt{7 + 9}$ $r = \sqrt{16}$ $r = 4$
• Calculate $r^2$:
  • Why this step? The standard equation of a circle uses $r^2$, so it's efficient to calculate this value directly to simplify the final substitution.
  • $r^2 = 4^2 = 16$.

Step 3: Write the Equation of the Circle

We now have all the necessary components for the circle's equation:

• Center $(h,k) = (0,3)$

• Radius squared $r^2 = 16$

• Why this step? This is the final step where we assemble all the calculated information into the required equation using the standard form of a circle's equation.

• Apply the Standard Equation of a Circle: $(x-h)^2 + (y-k)^2 = r^2$ Substitute the values: $(x-0)^2 + (y-3)^2 = 16$ $x^2 + (y^2 - 2 \cdot y \cdot 3 + 3^2) = 16$ $x^2 + y^2 - 6y + 9 = 16$

• Rearrange into General Form:

  • Why this step? The options provided are typically in the general form of a circle's equation ($x^2 + y^2 + Dx + Ey + F = 0$), so we rearrange our equation to match this format for easy comparison. $x^2 + y^2 - 6y + 9 - 16 = 0$ $x^2 + y^2 - 6y - 7 = 0$

Conclusion and Matching with Options

The equation of the circle is $x^2 + y^2 - 6y - 7 = 0$. Comparing this with the given 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) $x^2 + y^2 - 6y + 5 = 0$

Our calculated equation matches option (A).

---

Summary and Key Takeaway

This problem effectively tests your understanding of both ellipses and circles, requiring you to bridge concepts between the two conic sections. The methodical approach involves:

1. Extracting Ellipse Parameters: Correctly identifying $a^2$, $b^2$, and the orientation of the major axis from the given ellipse equation.
2. Calculating Eccentricity: Using the appropriate formula for eccentricity based on the major axis orientation.
3. Locating Foci: Determining the exact coordinates of the foci, which serve as points on the circle.
4. Determining Circle's Radius: Using the distance formula between the given circle's center and one of the foci.
5. Formulating Circle's Equation: Substituting the center and radius into the standard equation of a circle and simplifying.

Common Mistakes to Avoid:

• Confusing $a$ and $b$: Always remember that $a$ (or the larger denominator) corresponds to the semi-major axis, and $b$ (or the smaller denominator) to the semi-minor axis. This dictates the orientation and the correct eccentricity/foci formulas.
• Incorrect eccentricity formula: Ensure you use $e = \sqrt{1 - \frac{(\text{minor axis})^2}{(\text{major axis})^2}}$. If $a$ is the semi-major axis, use $a^2$ in the denominator; if $b$ is the semi-major axis, use $b^2$ in the denominator.
• Incorrect foci coordinates: If the major axis is on the x-axis, foci are $(\pm ae, 0)$. If on the y-axis, they are $(0, \pm ae)$ (assuming $a$ is the semi-major axis).
• Arithmetic and Algebraic Errors: Be meticulous with calculations involving squares, square roots, and expanding binomials like $(y-3)^2 = y^2 - 6y + 9$. Also, pay close attention to signs when rearranging equations ($9 - 16 = -7$).

By systematically following these steps and being mindful of potential pitfalls, you can confidently solve similar problems in JEE Mathematics.