# 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?

Nov 25, 2016

Yes they do, because the distance between the two circle centres $\mathbb{C} ' = 3$ is lower than the sum of both radii $R + R ' = 4$

#### Explanation:

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