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Q.

The line x+2y+6=0 cuts the coordinate axes at P and Q and a line perpendicular to it meet the axes in R and S. The equation to the locus of the point of intersection of the lines PS and QR is

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a

x2+y2+6x+3y=0

b

x2+y2−6x+3y=0

c

x2+y2−3x−3y=0

d

x2+y2−6x−6y=0

answer is A.

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Detailed Solution

Locus is the equation of circle with PQ as diametreClearly P(−6,0)&Q(0,−3) (x+6)x+y(y+3)=0⇒x2+y2+6x+3y=0

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