KEAM 2015 Math Paper

The graph of the both circles is shown below 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 $(x-1)^{2}+(y-1)^{2}=1$ $\Rightarrow\;\; (x-1)^{2}=\frac{1}{2} \Rightarrow x-1=\pm \frac{1}{\sqrt{2}}$ $\Rightarrow\;\; x=1+\frac{1}{\sqrt{2}}=\frac{\sqrt{2}+1}{\sqrt{2}}$ [we take positive sign due to 1st quadrant] $\therefore$ Required point $=\left(\frac{\sqrt{2}+1}{\sqrt{2}}, \frac{\sqrt{2}+1}{\sqrt{2}}\right)$