The equation of the circle passing through the foci of the ellipse $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$ 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.${x^2} + {y^2} - 6y + 5 = 0$

Question

The equation of the circle passing through the foci of the ellipse $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$ 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.${x^2} + {y^2} - 6y + 5 = 0$

Vedantu Content Team · Accepted Answer

Hint-In this question we have to find the equation of the circle passing through the foci of the ellipse $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$ but one must know briefly that ellipse is a plane curve surrounding two focal points and ellipse is determined by its foci as such, it generalises a circle but if you can use the lengths of the major and the minor axes to find its coordinates.

Complete step-by-step answer:

Before we go for the solution of this sum we must know the meaning of ellipse and foci of ellipse so in mathematics, ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal point is a constant. As such, it generalizes a circle, which is the special type of ellipse in which two focal points are the same and we should know that an ellipse is determined by its foci. But if you determine you can use the lengths of the major and the minor axes to find its coordinates. Now, the solution of the question is:
Coordinates of Foci are $( \pm ae,0)$
$a = 4,b = 3,e = \sqrt {1 - \dfrac{9}{{16}}} = \dfrac{{\sqrt 7 }}{4}$
Radius of the circle = $\sqrt {{{(ae)}^2} + {3^2}} $ =$\sqrt {16} = 4$
Equation of circle:
$
{\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16 \\
\Rightarrow {x^2} + {y^2} - 6y - 7 = 0 \\
$
Hence, option A is correct.
Note-In this question it is to be noted that for finding the equation of the circle passing through the foci of ellipse one should know the meaning of ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal point is a constant. As such it generalizes a circle. an ellipse is determined by its foci but if you determine you can use the lengths of the major and minor axes to find its coordinates. It should be noticed that I have used this definition a lot of times this is because one must clear his basics before moving on. Now, for a solution I have simply used the given values in the question to calculate the equation and simplify it.

The equation of the circle passing through the foci of the ellipse $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$ 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.${x^2} + {y^2} - 6y + 5 = 0$

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The equation of the circle passing through the foci of the ellipse $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$ 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.${x^2} + {y^2} - 6y + 5 = 0$

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The equation of the circle passing through the foci of the ellipse $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$ 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.${x^2} + {y^2} - 6y + 5 = 0$