Let S1:x2+y2−8x=0 S2:x2+y2−5x+7y=0 ‌S3:x2+y2−5x+7y−1=0 ‌S4:x2+y2−8x+7y−2=0 Now, at (5,−7) ‌S1=25+49−40=34>0 ‌S2=25+49−25−49=0 ‌S3=25+49−25−49−1=−1<0 ‌S4=25+49−40−49−2=−17<0 ∴(5,−7) lies outside the circle S1.