Question

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

• A. (2, -5)
• B. (5, -2)
• C. (-2, 5)
• D. (-5, 2)

Answer 1

Answer: (B)

Let the equation of circle be
$(x-3{)}^2+(y-0{)}^2+\lambda y=0$

As it passes through $(1,-2)$
$\therefore (1-3{)}^2+(-2{)}^2+\lambda (-2)=0$
$\Rightarrow 4+4-2\lambda =0\Rightarrow \lambda =4$
$\therefore$ Equation of circle is
$(x-3{)}^2+y^2+4y=0$
By hit and trial method, we see that point $(5,-2)$ satisfies equation of circle.