Question

lf $(1,2),(3,4)$ are limiting points and $x^2+y^2-x+ky=0$ is one circle of a coaxial system then $k=$

• A. $3$
• B. $-3$
• C. $-9$
• D. $9$

Answer 1

The correct option is B $-3$

$e{q}^n$ of line passing through limiting point is
$y-2=(x-1)$
$y-x-1=0--\left(1\right)$
So, centres of all circles lie on $y=x$ axis
So, $x^2+y^2-x+ky=0$
centre $(\frac{1}{2},\frac{-k}{2})$
$\frac{-k}{2}-\frac{1}{2}-1=0$
$\frac{k}{2}=\frac{-3}{2}$
$\Rightarrow k=-3$