Given equation of circle x2+y2+4x−10y−7=0 with centre =(−2,5)=C and radius =√4+25+7=√36=6
Now, 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