From both the equations of circles, we observe that the centre of the circles lie on the straight line y=x So, the nearest point on the first circle w.r.t. second circle will be obtained on putting y=x in equation of circle (x−1)2+(y−1)2=1
⇒‌‌(x−1)2+(x−1)2=1‌‌[∵y=x]
⇒‌‌(x−1)2=‌
1
2
⇒x−1=±‌
1
√2
⇒‌‌x=1+‌
1
√2
=‌
√2+1
√2
[we take positive sign due to 1st quadrant] ∴ Required point =(‌