Question

The value of k for which the circles $x^2+y^2-3x+1$ k y - 5 = 0 and $4x^2+4y^2-12x-y-9=0$ become concentric is

• A. $-\frac{1}{8}$
• B. $\frac{1}{8}$
• C. $\frac{1}{4}$
• D. $-\frac{1}{4}$

Answer 1

The correct option is D $-\frac{1}{4}$

The given equation of circles are,

$\Rightarrow$ $x^2+y^2-3x+1ky-5=0$ --- ( 1 )

$\Rightarrow$ $4x^2+4y^2-12x-y-9=0$ --- ( 2 )

Comparing the equation ( 1 ) with $x^2+y^2+2gx+2fy+c=0$ we get,

$\Rightarrow$ $2g=-3$

$\therefore$ $g=\frac{-3}{2}$

$\Rightarrow$ $2f=k$

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

We know, $C=(-g,-f)$

$\therefore$ $C_1=(\frac{3}{2},\frac{-k}{2})$ ---- ( 3 )

Now,

Dividing equation ( 2 ) by ( 4 ) we get,

$\Rightarrow$ $x^2+y^2-3x-\frac{y}{4}-\frac{9}{4}=0$

Comparing the above equation with $x^2+y^2+2gx+2fy+c=0$ we get,

$\Rightarrow$ $2g=-3$

$\therefore$ $g=\frac{-3}{2}$

$\Rightarrow$ $2f=-\frac{1}{4}$

$\Rightarrow$ $f=\frac{-1}{8}$

$C=(-g,-f)$

$\therefore$ $C_2=(\frac{3}{2},\frac{1}{8})$ ---- ( 4 )

The circles are concentric and we know that centers of concentric circles are equal.

$\therefore$ $C_1=C_2$

$\Rightarrow$ $(\frac{3}{2},\frac{-k}{2})$ $=(\frac{3}{2},\frac{1}{8})$ [ From ( 3 ) and ( 4 ) ]

$\therefore$ $\frac{-k}{2}=\frac{1}{8}$

$\therefore$ $k=\frac{-1}{4}$