Question

# The equation of the circle whose diameter is the common chord of the circles x2+y2+3x+2y+1=0 and x2+y2+3x+4y+2=0 is

A
x2+y2+3x+y+5=0
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B
x2+y2+x+3y+7=0
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C
x2+y2+2x+3y+1=0
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D
2(x2+y2)+6x+2y+1=0
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Solution

## The correct option is D 2(x2+y2)+6x+2y+1=0REF.Image.Given circles(1) x2+y2+3x+2y+1=0(2) x2+y2+3x+4y+2=0then common chord ⇒ there is 2 common pointsof intersection for (1) and (2)So (2) - (1) ⇒2y=−1or y=−12Now, ∵ the required circle has this common chordas its diameter the eqn of circle shouldbe such that it has 'y' common pt of intersectingwith (1) and (2)So (3) should be each thatx2+y2+ax+by+c=0(3) - (1) ⇒y=−12 common pt of intersectionand (3) - (2) ⇒y=−12 (see fig a,b)from given options(D) 2(x2+y2)+3x+2y+1=0or x2+y2+3x+y+12=0 has(D) −1⇒y=−1/2(D) −2⇒−3y−3/2=0 or y=−12For all other options we do not get common pt of intersection

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