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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

Equation of the circle with centre on the $y$-axis and passing through the origin and the point $(2,3)$ is

$\begin{array}{1 1}x^2+y^2+13y=0\\3x^2+3y^2+13x+3=0\\6x^2+6y^2-13x=0\\x^2+y^2+13x+3=0\end{array} $

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A)
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  • General equation of a parabola whose centre is on the $y$ axis is $(x-0)^2+(y-k)^2=k^2$ (ie) $x^2+(y-k)^2=k^2$
Answer : $6x^2+6y^2-13x=0$
The given circle passes through the point $(2,3)$ is
$x^2+(y-k)^2=k^2$
(ie) $2^2+(3-k)^2=k^2$
On expanding we get,
$4+9-6k+k^2=k^2$
$6k=13$
$k=\large\frac{13}{6}$
Substituting the value of $k$ we get,
$x^2+(y-\large\frac{13}{6})^2=(\large\frac{13}{6})^2$
On simplifying we get,
$6x^2+6y^2-13x=0$
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