# Circle A has a center at (-4 ,6 ) and a radius of 4 . Circle B has a center at (1 ,1 ) and a radius of 2 . Do the circles overlap? If not what is the smallest distance between them?

Mar 16, 2016

The circles do not overlap.
Smallest distance between them is 4

#### Explanation:

If you find the distance between centres then directly compare it to the sum of the radii you can determine if they do overlap or not.

Let the line length be $L$
Let the radius ${r}_{1} = 4$
Let the radius ${r}_{2} = 2$

Using Pythagoras

Then $L = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$L = \sqrt{{\left(1 - \left(- 4\right)\right)}^{2} + {\left(1 - 6\right)}^{2}}$

$L = \sqrt{25 + 25}$

$L = 10$

The sum of the radii is ${r}_{1} + {r}_{2} = 4 + 2 = 6$

Thus ${r}_{1} + {r}_{2} < L$ so the circles do not overlap

Distance between $\to L - {r}_{1} - {r}_{2} = 4$

Mar 16, 2016

no overlap , distance ≈ 1.071

#### Explanation:

First step is to calculate the distance between the centres using the $\textcolor{b l u e}{\text{ distance formula }}$

 d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2

where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points }$

let $\left({x}_{1} , {y}_{1}\right) = \left(- 4 , 6\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 1\right)$

hence :  d = sqrt((1-(-4))^2 + (1-6)^2) = sqrt(25+25 )= sqrt50 ≈ 7.071

now : radius of A + radius of B = 4+2 =6

since sum of radii < distance between centres → no overlap

distance between circles = 7.071 - 6 = 1.071