Question

The circle passing through (1, -2) and touching the axis of x 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)

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

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.