Question

Find the equation of circle which passes through (0,0),(2,0),(0,3).

Answer 1

Answer:

Equation of the required circle is

$x^2+y^2-2x-3y=0$

Step-by-step explanation:

Formula used:

General equation of circle is

$x^2+y^2+2gx+2fy+c=0$

Let the equation of the required circle be

$x^2+y^2+2gx+2fy+c=0$.......(1)

Circle (1) passes through (0,0),

$0^2+0^2+2g(0)+2f(0)+c=0$

c=0

Circle (1) passes through (2,0),

$2^2+0^2+2g(2)+2f(0)+0=0$

$4+4g=0$

$4g=-4$

$g=-1$

Circle (1) passes through (0,3),

$0^2+3^2+2g(0)+2f(3)+0=0$

$9+6f=0$

$6f=-9$

$f=\frac{-9}{6}$

$f=\frac{-3}{2}$

Now equation (1) becomes,

$x^2+y^2+2(-1)x+2(\frac{-3}{2})y+0=0$

$x^2+y^2-2x-3y=0$