Question

The radius of the circle passing through the foci of the ellipse \(\left(\mathrm{x}^{2} / 16\right)+\left(\mathrm{y}^{2} / 9\right)=1\) and having its centre \((0,3)\) is \(\ldots \ldots\) (a) 4 (b) 3 (c) \(\sqrt{12}\) (d) \((7 / 2)\)

Short answer

The radius of the circle passing through the foci of the ellipse \(\left(\frac{x^2}{16}\right)+\left(\frac{y^2}{9}\right)=1\) and having its center at \((0, 3)\) is \(4\).

Step-by-step solution

Find the foci of the ellipse

The equation of the ellipse is given by \(\left(\mathrm{x}^{2} / 16\right)+\left(\mathrm{y}^{2} / 9\right)=1\). This is in the standard form for an ellipse with horizontal major axis: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 = 16\) and \(b^2 = 9\). The distance between the center of the ellipse and each focus is given by \(c = \sqrt{a^2 - b^2}\). Calculate the value of \(c\): \(c = \sqrt{16 - 9}\) \(c = \sqrt{7}\) Now, we can find the coordinates of the foci. Since the ellipse has a horizontal major axis and is centered at the origin, the foci are located at \((-c, 0)\) and \((c, 0)\), or \((-\sqrt{7}, 0)\) and \((\sqrt{7}, 0)\).

Find the radius of the circle

The circle has its center at (0, 3), and we just found the coordinates of the foci to be \((-\sqrt{7}, 0)\) and \((\sqrt{7}, 0)\). Since the circle passes through both foci, the distance from its center to each focus is the radius. We will use the distance formula to find this radius: \(r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) Calculate the distance between the center of the circle and one of the foci, for example, \((\sqrt{7}, 0)\): \(r = \sqrt{(0 - \sqrt{7})^2 + (3 - 0)^2}\) \(r = \sqrt{7 + 9}\) \(r = \sqrt{16}\) \(r = 4\) So, the radius of the circle is 4, which corresponds to choice (a). #Answer#: (a) 4

Key concepts

Ellipse Properties

An ellipse is a special kind of conic section that looks like a stretched circle. It has two main properties: a major axis and a minor axis. The longest diameter is the major axis, while the minor axis is the shortest. The standard equation for an ellipse with a horizontal major axis is given by:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]where:

• \(a\) is the semi-major axis length.
• \(b\) is the semi-minor axis length.
• If \(a > b\), the major axis is horizontal; if \(b > a\), it’s vertical.

A unique feature of the ellipse is its foci (plural of focus). These are two fixed points inside the ellipse. For the given ellipse, \(a^2 = 16\) and \(b^2 = 9\). The distance \(c\) from the center to each focus is calculated by the formula \(c = \sqrt{a^2 - b^2}\). The foci of this ellipse are thus located at \((-\sqrt{7}, 0)\) and \((\sqrt{7}, 0)\). These points are crucial for understanding various ellipse problems and their solutions.

Distance Formula

The distance formula is vital for finding the length of a line segment between two points in the plane. If you want to calculate how far apart two points, say \((x_1, y_1)\) and \((x_2, y_2)\), are, you use the formula:\[\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This equation uses basic principles from the Pythagorean theorem, and it's a fundamental tool in analytic geometry. For our exercise, this formula helps determine the radius of the circle by measuring the distance between the circle’s center at \((0, 3)\) and its foci \((\sqrt{7}, 0)\). We substitute these coordinates into the formula to compute:

• \((x_1, y_1) = (0, 3)\)
• \((x_2, y_2) = (\sqrt{7}, 0)\)

The result from the formula tells us the radius is 4 units. This illustrates how essential the distance formula is in solving geometric problems involving points and lines.

Circle Properties

A circle is a simple geometric shape that is defined by all the points that are a fixed distance from a center. This fixed distance is called the radius. The equation of a circle centered at point \((h, k)\) with radius \(r\) is:\[(x - h)^2 + (y - k)^2 = r^2\]where:

• \((h, k)\) is the center of the circle.
• \(r\) is the radius, the distance from the center to any point on the circle.

In the given problem, the circle is described to pass through foci of an ellipse and its center is given as \((0, 3)\).
Using properties of circles, if any point like a focus lies on a circle's circumference, the distance from the center to that point equals the radius of the circle.
By using the distance formula, we calculated that this radius is 4, confirming one of the multiple-choice answers. Understanding these circle properties allows better comprehension of circle equations and geometry.