Consider a circle $C_1$with equation $(\mathrm{x}-1{)}^2+{\mathrm{y}}^2=1$ and a shrinking circle $C_2$ with radius $r$ and centre at the origin. $P$ is the point $\left(0,r\right)$,"Q" is the upper point of inter section (above $x-$axis) of the two circles and "R" is the point of intersection of the line PQ with $x-$axis. then $x-$coordinate of the point $R$, as C₂ shrinks, $\text{ (i.e., }r\to 0\text{ )? }$

Suppose that the point $R\left(x,0\right)$ and the points $P\left(0,r\right)$ 

The point of intersection of both circles is $Q$

             $\begin{array}{rcl}{\left(x-1\right)}^2+y^2& =& 1\\ x^2+y^2& =& r^2\end{array}$

subtracting one from the other 

            $\begin{array}{rcl}{\left(x-1\right)}^2-x^2& =& 1-r^2\\ -2x+1& =& 1-r^2\\ x& =& \frac{r^2}{2}\end{array}$

Hence, the point is $Q\left(\frac{r^2}{2},\sqrt{r^2-\frac{r^4}{4}}\right)$

P, Q, R are collinear points

Slope of $PR$ is same as the slope of $PQ$

it implies that 

            $\begin{array}{rcl}-\frac{r}{x}& =& \frac{r\sqrt{\left(1-{\displaystyle \frac{r^2}{4}}\right)}-r}{{\displaystyle \frac{r^2}{2}}}\\ -\frac{1}{x}& =& \frac{2}{r^2}\left(\sqrt{\left(1-\frac{r^2}{4}\right)}-1\right)\\ x& =& \frac{r^2}{2-\sqrt{4-r^2}}\end{array}$

Hence, 

                  $\begin{array}{rcl}\underset{r\to 0}{Lim} x& =& \underset{r\to 0}{Lim} \frac{r^2}{2-\sqrt{4-r^2}}\\ & =& \underset{r\to 0}{Lim} \frac{2r}{{\displaystyle \frac{1}{2\sqrt{4-r^2}}}\left(2r\right)} LH rule\\ & =& \underset{r\to 0}{Lim} 2\sqrt{4-r^2}\\ & =& 4\end{array}$

Therefore as $r\to 0$ the value of $x$ tends to $\boxed{4}$