# Circle A has a radius of 2  and a center at (5 ,1 ). Circle B has a radius of 1  and a center at (3 ,2 ). If circle B is translated by <-2 ,6 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Feb 15, 2018

The circle B does not overlap circle A.
The minimum distance between points on both circles is 3 units

#### Explanation:

Circle A:
radius ${r}_{1} = 2$
center $A \equiv \left(5 , 1\right)$
Circle B:
radius ${r}_{2} = 1$
center $B \equiv \left(3 , 2\right)$
translation of B $B ' \equiv < - 2 , 6 >$
new center $\left(3 - 2 , 2 + 6\right) = \left(5 , 8\right)$
Distance between center of A and new center of B' is
$\sqrt{{\left(5 - 5\right)}^{2} + {\left(8 - 1\right)}^{2}} = \sqrt{{0}^{2} + {7}^{2}} = {\sqrt{7}}^{2} = 7$
$A B ' = 7$
radius of A ${r}_{1} = 2$
radius of B ${r}_{2} = 1$

Sum of the radii ${r}_{1} + {r}_{2} = 2 + 1 = 3$

Sum of the radii $<$ Distance between centers A and B.
Hence, they do d
not overlap.
The difference between
the distance between the centers'an
represents the shortest distance between the two circles

separation $= 7 - 3 = 4$

The two circles are separated by 3 units

Feb 15, 2018

$\text{no overlap } \approx 5.062$

#### Explanation:

$\text{what we have to do here is "color(blue)"compare ""the distance (d)}$
$\text{between the centres to the "color(blue)"sum of radii}$

• " if sum of radii">d" then circles overlap"

• " if sum of radii"< d" then no overlap"

$\text{before calculating d we require to find the centre of}$
$\text{B under the given translation}$

$\text{under the translation } < - 2 , 6 >$

$\left(3 , 2\right) \to \left(3 - 2 , 2 + 6\right) \to \left(1 , 8\right) \leftarrow \textcolor{red}{\text{new centre of B}}$

$\text{to calculate d use the "color(blue)"distance formula}$

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

$\text{let "(x_1,y_1)=(5,1)" and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 8\right)$

$d = \sqrt{{\left(1 - 5\right)}^{2} + {\left(8 - 1\right)}^{2}} = \sqrt{16 + 49} = \sqrt{65} \approx 8.062$

$\text{sum of radii } = 2 + 1 = 3$

$\text{since sum of radii"< d" then no overlap}$

$\text{min. distance "=d-" sum of radii}$

$\textcolor{w h i t e}{\times \times \times \times \times} = 8.062 - 3 = 5.062$
graph{((x-5)^2+(y-1)^2-4)((x-1)^2+(y-8)^2-1)=0 [-20, 20, -10, 10]}