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
- 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 & 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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