Question

If a circle passes through the points where the lines $3kx-2y-1=0$ and $4x-3y+2=0$ meet the coordinate axes then $k=$

• A. $1$
• B. $-1$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (C)

$\frac{1}{2}$

The line $3kx-2y-1=0$ meets the x-axis and y-axis at

$A(\frac{1}{3k},0)$ and $B(0,-\frac{1}{2})$ respectively

and the line $4x-3y+2=0$ cuts x-axis and y-axis at

$C(-\frac{1}{2},0)$ and $D(0,\frac{2}{3})$ respectively.

Since the four points are concyclic,

$\therefore OA.OC=OB.OD$

$\Rightarrow \frac{1}{3\left|k\right|}.\frac{1}{2}=\frac{1}{2}.\frac{2}{3}\Rightarrow \left|k\right|=\frac{1}{2}$

According to the given geometrical position $k$ must be positive

$\therefore k=\frac{1}{2}$