Problem 80 - A Complete Resource Book in Mathematics for JEE Main 2017

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: (A) 4 unit (B) 3 unit (C) \(\sqrt{12}\) unit (D) \(\frac{7}{2}\) unit

Step-by-Step Solution

Answer

The radius of the circle is 4 units.

Step 1: Identify the Foci of the Ellipse

The given ellipse is \(\frac{x^2}{16} + \frac{y^2}{9} = 1\). The equation is in the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 = 16\) and \(b^2 = 9\). Thus, \(a = 4\) and \(b = 3\). For an ellipse with \(a > b\), the distance from the center to each focus \(c\) is given by \(c = \sqrt{a^2 - b^2}\). Calculating \(c\): \(c = \sqrt{16 - 9} = \sqrt{7}\). Hence, the foci of the ellipse are at \((\pm \sqrt{7}, 0)\).

Step 2: Determine the Center of the Circle

The center of the circle is given in the problem statement as \((0, 3)\).

Step 3: Calculate the Distance from the Circle Center to Foci

To find the radius of the circle, calculate the distance between its center \((0, 3)\) and one of the ellipse's foci \((\sqrt{7}, 0)\). The distance formula is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Using points \((0, 3)\) and \((\sqrt{7}, 0)\), we calculate: \(d = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2} = \sqrt{7 + 9} = \sqrt{16} = 4\).

Step 4: Conclude the Radius of the Circle

Since the distance calculated is the radius of the circle, the radius is 4 units.

Key Concepts

Circle

A circle is a simple and well-known geometric shape. It consists of all points in a plane that are at an equal distance from a fixed point, known as the center. This constant distance is called the radius of the circle. Let's break down the key points of a circle:

• Center: The fixed point inside the circle from which every point on the perimeter is equidistant.
• Radius: The distance from the center to any point on the circle. In our problem, this is what we are trying to calculate using the distance from the center of the circle to the foci of the ellipse.

The equation of a circle with center \( (h, k) \) and radius \( r \) is given by \( (x-h)^2 + (y-k)^2 = r^2 \). This formula helps us understand how circles can be described in coordinate geometry, making it easier to visualize and solve related problems.

Foci of ellipse

In geometry, an ellipse is an elongated circle. Important to the definition of an ellipse are its two foci (plural of focus). Unlike a circle, which has a single center, an ellipse's properties revolve around these two foci. Here are some key characteristics:

• Foci: For the given ellipse \((\frac{x^2}{16} + \frac{y^2}{9} = 1)\), the foci are calculated using the formula \(c = \sqrt{a^2 - b^2}\), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. Thus, \(c = \sqrt{16 - 9} = \sqrt{7}\), and the foci are located at \( (\pm \sqrt{7}, 0) \).
• Role of Foci: The sum of the distances from any point on the ellipse to the two foci is constant. This unique property helps in understanding other shapes, such as the circle passing through these foci in our problem.

Distance formula

The distance formula is a tool used in geometry to determine the distance between two points in a plane. It's derived from the Pythagorean theorem and is expressed as \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Here's how it works:

• Usage: To find the radius of the circle in our problem, we use the distance formula to calculate the distance between the center of the circle \( (0, 3) \) and one of the ellipse's foci \( (\sqrt{7}, 0) \).
• Calculation: By plugging in the values, \(d = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2} = \sqrt{7 + 9} = \sqrt{16} = 4\), we find that the radius of the circle is 4 units.

Using this formula is a straightforward process, as it simply applies the basic principles of geometry to calculate distances accurately between two coordinates.