Question

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

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

Answer 1

The correct option is C $(5,-2)$

$\therefore$ Equation of required circle is
$(x-3{)}^2+y^2+\lambda y=0$
As, $(1,-2)$ satisfy it, so $4+4=2\lambda$
$\Rightarrow \lambda =4$
$\therefore$ The equation of circle is

$(x-3{)}^2+y^2+4y=0$

$x^2+y^2-6x+4y+9=0$

By hit and trial method, for the point $(5,-2)$

Put $x=5$ and $y=-2$ in the above equation of circle

$(5{)}^2+(-2{)}^2-6\left(5\right)+4(-2)+9$

$=25+4-30-8+9$

$=38-38=0$

Clearly, $(5,-2)$ satisfies the equation of the given circle. so the circle passes through $(5,-2)$