2007
Problem - 4755
Find the point on the circle $(x − 5)^2 + (y − 4)^2 = 4$ which is closest to the circle $(x − 1)^2 + (y − 1)^2 = 1$.
The center of the first circle $O_1$ is $(5, 4)$ and the center of the second center $O_2$ is $(1,1)$. Their radii are $2$ and $1$, respectively. The distance between $O_1$ and $O_2$ is $\sqrt{(5-1)^2 + (4-1)^2}=5$ which is greater than the sum of their radii which equals $2+1=3$. Therefore, the desired point, $P$, is on the line $O_1O_2$ and satisfying $O_1P : PO_2 = 2 : 3$. Therefore, we have $$P (x, y) = \left(\frac{2\times 1 + 3\times 5}{2+3}, \frac{2\times 1 + 3\times 4}{2+3}\right)=\boxed{\left(\frac{17}{5}, \frac{14}{5}\right)}$$