Locus of point of intersection of perpendicular tangents drawn one to each of the circles x2+y2−2x−4y−3=0;x2+y2−2x−4y+1=0 is

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x2+y2−2x−4y−7=0

x2+y2+2x+4y−7=0

x2+y2+2x−4y−7=0

x2+y2−2x+4y+7=0

answer is A.

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Detailed Solution

c1=c2=(1,2) r1=8,r2=2 Required locus is (x−1)2+(y−2)2=r12+r22 ⇒x2+y2−2x−4y−7=0

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