Question

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+3x+2y+1=0$ and $x^2+y^2+3x+4y+2=0$ is

• A. $x^2+y^2+3x+y+5=0$
• B. $x^2+y^2+x+3y+7=0$
• C. $x^2+y^2+2x+3y+1=0$
• D. $2(x^2+y^2)+6x+2y+1=0$

Answer 1

The correct option is D $2(x^2+y^2)+6x+2y+1=0$

REF.Image.

Given circles

(1) $x^2+y^2+3x+2y+1=0$

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

then common chord $\Rightarrow$ there is 2 common points

of intersection for (1) and (2)

So (2) - (1) $\Rightarrow 2y=-1$

or $y=\frac{-1}{2}$

Now, $\because$ the required circle has this common chord

as its diameter the eqn of circle should

be such that it has 'y' common pt of intersecting

with (1) and (2)

So (3) should be each that

$x^2+y^2+ax+by+c=0$

(3) - (1) $\Rightarrow y=\frac{-1}{2}$ common pt of intersection

and (3) - (2) $\Rightarrow y=\frac{-1}{2}$ (see fig a,b)

from given options

(D) $2(x^2+y^2)+3x+2y+1=0$

or $x^2+y^2+3x+y+\frac{1}{2}=0$ has

(D) $-1\Rightarrow y=-1/2$

(D) $-2\Rightarrow -3y-3/2=0$ or $y=\frac{-1}{2}$

For all other options we do not get common pt of intersection