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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
Conic Sections
Solution:
Let the centre be $\left(0, k\right)$.
The equation of circle be
$x^{2} + y^{2} - 2\left(0\right)x - 2ky + c = 0$
$\Rightarrow x^{2}+y^{2}-2ky + c = 0\, \ldots\left(i\right)$
Since, it passes through $\left(0,0\right)$ and $\left(2,3\right)$.
$\therefore \, 0^{2} + 0^{2} - 2k\left(0\right) + c = 0$
$\Rightarrow c=0$
and $\left(2\right)^{2} + \left(3\right)^{2} - 2k\left(3\right) + c = 0$
$\Rightarrow 13-6k + c = 0$
$\Rightarrow 13- 6k+0-0 = 0$
$\Rightarrow k=\frac{13}{6}$
On putting the values of $k$ and $c$ in $\left(i\right)$, we get
$x^{2}+y^{2}-2\left(\frac{13}{6}\right)y+0=0$
$\Rightarrow 6x^{2} + 6y^{2}- 26y = 0$