# Circle A has a radius of 1  and a center at (3 ,3 ). Circle B has a radius of 3  and a center at (6 ,4 ). If circle B is translated by <-3 ,4 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Yes they do, because the distance between the two circle centres $\mathbb{C} ' = 3$ is lower than the sum of both radii $R + R ' = 4$
The circle A has equation ${\left(x - 3\right)}^{2} + {\left(y - 3\right)}^{2} = 1$ with centre C(3;3) and radius $R = 1$, whereas the circle B has equation ${\left(x - 6\right)}^{2} + {\left(y - 4\right)}^{2} = 9$ with radius $R ' = 3$.
If we translate the second circle by the vector $\left(- 3 , 4\right)$ the new equation of B circle is ${\left(x - 3\right)}^{2} + {y}^{2} = 9$ and centre C'(3;0) and $R ' = 3$
The distance between the two centres is $\mathbb{C} ' = \sqrt[2]{{\left(3 - 3\right)}^{2} + {\left(3 - 0\right)}^{2}} = 3$. As the $\mathbb{C} '$ is less than the sum of the two radius we can deduce that the two circles overlap