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. ${x^2} + {y^2} - 6y - 5 = 0$
C. ${x^2} + {y^2} - 6y + 5 = 0$
D. ${x^2} + {y^2} - 6y - 7 = 0$

Answer 1

Hint: To solve this question, we should remember the basic points of the standard equation of circle and as well as ellipse. The foci of the ellipse will be $\left( { \pm ae,0} \right)$ where a is the x coordinate of the ellipse and e is the eccentricity. The distance between the foci and the centre of the circle will be the radius of that circle.

Complete step-by-step answer:
Given that,
Equation of ellipse = $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$ ………… (i)
Centre of circle = $\left( {0,3} \right)$
The standard equation of ellipse is given by,
$ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
Comparing this with equation (i), we will get
a = 4 and b = 3.
Here, we can see that a > b,
So, the foci of the ellipse will be $\left( { \pm ae,0} \right)$
We know that,
Eccentricity, $e = \sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}} $
Putting values of a and b, we will get
\[
   \Rightarrow e = \sqrt {1 - \dfrac{{{3^2}}}{{{4^2}}}} \\
   \Rightarrow e = \sqrt {1 - \dfrac{9}{{16}}} \\
   \Rightarrow e = \dfrac{{\sqrt 7 }}{4} \\
\]
Therefore,
Foci = $\left( { \pm \sqrt 7 ,0} \right)$
As according to the question, the circle is passing through the foci of the ellipse, i.e. $\left( { \pm \sqrt 7 ,0} \right)$ and we have already given the centre of the circle as (0,3).
Therefore,
Radius of circle = distance between $\left( {\sqrt 7 ,0} \right)$ and $\left( {0,3} \right)$.
Radius, r = $\sqrt {{{\left( {\sqrt 7 - 0} \right)}^2} + {{\left( {0 - 3} \right)}^2}} $
r = $\sqrt {7 + 9} = \sqrt {16} $
r = $4$.
We know that, the general equation of a circle is given by,
$ \Rightarrow {\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, where (h,k) is the centre of the circle.
Putting all the values here, we will get
\[
   \Rightarrow {\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = {4^2} \\
   \Rightarrow {x^2} + {y^2} - 6y + 9 = 16 \\
   \Rightarrow {x^2} + {y^2} - 6y - 7 = 0 \\
\]
Hence, the required equation of the circle is \[{x^2} + {y^2} - 6y - 7 = 0\].
Therefore, the correct answer is option (A).

Note: Whenever we ask such type of questions, first, we have to know some basic points of the ellipse. The eccentricity of an ellipse is also given by, $\dfrac{c}{a}$ and $c = \sqrt {{a^2} - {b^2}} $. When c = 0, both foci merge together with the centre of the ellipse and ${a^2} = {b^2}$, i.e. a = b, and so the ellipse becomes a circle.