Question

find the eq of circle passing through the point (2,3) whose centre lies on y axis and passes throught the origin

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that centre of circle be (0, k) and radius of circle be r units.

So, equation of circle is given by

$\rm \: {(x - 0)}^{2} + {(y - k)}^{2} = {r}^{2}$

$\rm \: {x}^{2} + {(y - k)}^{2} = {r}^{2} - - - (1)$

Now, Since circle (1) passes through (0, 0), we get

$\rm \: {0}^{2} + {(0 - k)}^{2} = {r}^{2}$

$\rm \: {(- k)}^{2} = {r}^{2}$

$\rm \: {k}^{2} = {r}^{2} - - - (2)$

Now, further given that circle (1), passes through (2, 3), so we get

$\rm \: {2}^{2} + {(3 - k)}^{2} = {r}^{2}$

$\rm \: 4 + 9 + {k}^{2} - 6k = {k}^{2}$

$\rm \: 13 - 6k = 0$

$\rm \: - 6k = - 13$

$\rm\implies \:k \: = \: \dfrac{13}{6}$

So, on substituting the value of k in equation (2), we get

$\rm\implies \: {r}^{2} \: = \: \dfrac{169}{36}$

So, on substituting the values of k and r in equation (1), we get

$\rm \: {x}^{2} + {\bigg(y - \dfrac{13}{6} \bigg) }^{2} = \dfrac{169}{36}$

$\rm \: {x}^{2} + {y}^{2} + \frac{169}{36} - 2 \times y \times \frac{13}{6} = \dfrac{169}{36}$

$\rm \: {x}^{2} + {y}^{2} - \frac{13y}{3} = 0$

$\bf\implies \: \: {3x}^{2} + {3y}^{2} - 13y = 0$

is the required equation of circle passing through the point (2,3) whose centre lies on y axis and passes throught the origin.