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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

The  the circle passing through the foci of the \(\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1\) and having centre at \((0,3) \) is

  1. A \({x^2} + {y^2} - 6y - 7 = 0\)
  2. B \(\;{x^2} + {y^2} - 6y + 7 = 0\)
  3. C \(\;{x^2} + {y^2} - 6y - 5 = 0\)
  4. D \(\;{x^2} + {y^2} - 6y + 5 = 0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \({x^2} + {y^2} - 6y - 7 = 0\)

Step-by-step Solution

Detailed explanation

\(a=4, b=3, e=\sqrt{1-\frac{9}{16}} \quad \Rightarrow \quad \frac{\sqrt{7}}{4}\) Foci is \((\pm \text { ae }, 0)\) \(\Rightarrow \quad(\pm \sqrt{7}, 0)\) \(r=\sqrt{(a e)^{2}+b^{2}}\) \(\sqrt{7+9}\) \(=4\) Now equation of circle is \((x-0)^{2}+(y-3)^{2}=16\)…
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