KEAM 2011 Math Paper

Given equation of circle
x2+y2+4x−10y−7=0
with centre =(−2,5)=C and
radius =√4+25+7=√36=6Now, PC=√(4+2)2+(−3−5)2
=√36+64=√100=10
So, the point lies outside the circle.
Now, maximum distance
=PQ=PC+QC=10+6=16
and minimum distance
=PS=PC−SC=10−6=4
∴ Sum of maximum and minimum distance
=16+4=20