Discussion Forum

The circle passing through (1,-2) and touching the axis of x at (3,0) also passes through the point 

Answer:

Explanation:

 Let the equation of circle be 

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

 As it passes  through  (1,-2)

 $\therefore$   $(1-3)^{2}+(-2)^{2}+\lambda (-2)=0$

 $\Rightarrow$    $ 4+4-2\lambda=0\Rightarrow\lambda=4$

 $\therefore$    $(x-3)^{2}+y^{2}+4y=0$   By hit and trial method, we see  that point(5,-2) satisfies equation of circle.