Three points on circle are (1,1),(−2,2),(2,−2) Let equation of circle be x2+y2+2gx+2fy+c=0 ∵ Above points lie on the circle ‌∴‌‌1+1+2g+2f+c=0 ‌⇒‌‌2g+2f+c+2=0‌‌...‌ (i) ‌ ‌(−2)2+(2)2+2(−2)g+2×2×f+c=0 ‌8−4g+4f+c=0 ‌‌‌4g−4f−c−8=0‌‌...‌ (ii) ‌ and (2)2+(−2)2+2×2×g+2(−2)f+c=0 4g−4f+c+8=0‌‌...‌ (iii) ‌ On solving Eqs. (i), (ii) and (iii), we get ‌c=−8,g=‌
3
2
,f=‌
3
2
‌∴‌ Centre of circle ‌≡(‌
−3
2
,‌
−3
2
) Now, perpendicular distance from centre of the circle to the line ‌=|‌
