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

• 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. $x^2+y^2-6y+5=0$

Answer 1

The Correct Option is A

To find the equation of the circle passing through the foci of the given ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) with the center at (0, 3), we need to follow these steps:

1. The given ellipse is of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where \(a^2 = 16\) and \(b^2 = 9\).
2. Since \(a^2 > b^2\), the ellipse has a horizontal major axis. The foci are calculated using the formula \(c = \sqrt{a^2 - b^2}\) where \(c\) is the distance from the center to each focus.
3. Calculate \(c\): 

\[c = \sqrt{16 - 9} = \sqrt{7}.\]

1.  Thus, the foci of the ellipse are at the points \((\pm \sqrt{7}, 0)\).
2. The center of the required circle is given as \((0, 3)\).
3. Let the radius of the circle be \(r\). Since the circle must pass through both foci \((\sqrt{7}, 0)\) and \((-\sqrt{7}, 0)\), we calculate the distance from these points to the center of the circle:
4. Consider the distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\): 

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.\]

1.  For \((0, 3)\) and \((\sqrt{7}, 0)\): 

\[d = \sqrt{(0 - \sqrt{7})^2 + (3 - 0)^2} = \sqrt{7 + 9} = \sqrt{16} = 4.\]

1. The radius \(r\) of the circle is thus 4.
2. The standard equation of a circle with center \((h, k)\) and radius \(r\) is: 

\[(x - h)^2 + (y - k)^2 = r^2.\]

1.  Substituting the values \(h = 0\), \(k = 3\), and \(r = 4\), we have: 

\[(x - 0)^2 + (y - 3)^2 = 16.\]

1.  Simplifying, this becomes: 

\[x^2 + (y^2 - 6y + 9) = 16.\]

1.  Which simplifies to: 

\[x^2 + y^2 - 6y - 7 = 0.\]

Therefore, the equation of the circle is \(x^2 + y^2 - 6y - 7 = 0\).