# Circle A has a center at (1 ,3 ) and a radius of 1 . Circle B has a center at (-2 ,-5 ) and a radius of 2 . Do the circles overlap? If not, what is the smallest distance between them?

May 20, 2017

The distance between the centres is given by:

$r = \sqrt{{\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {x}_{1}\right)}^{2}} = 8.54$ units

Since this is larger than 3 units, the sum of the radii, the circles do not overlap.

#### Explanation:

The sum of the two radii is $2 + 1 = 3$ units. If the centres are less than $3$ units apart, the circles overlap, if they are more than $3$ units apart then they do not.

The distance between the centres is given by:

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

$r = \sqrt{{\left(- 5 - 3\right)}^{2} + {\left(- 2 - 1\right)}^{2}} = \sqrt{{\left(- 8\right)}^{2} + {\left(- 3\right)}^{2}}$

$\therefore r = \sqrt{64 + 9} = \sqrt{73} = 8.54$ units

Since this is larger than 3 units, the circles do not overlap, and this is the distance between their centres.

May 20, 2017

$\text{no overlap, min. distance" } \approx 5.544$

#### Explanation:

What we have to do here is $\textcolor{b l u e}{\text{compare}}$ the distance ( d) between the centres of the circles to the $\textcolor{b l u e}{\text{sum of the radii}}$

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

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

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ are 2 coordinate points}$

$\text{the points are } \left({x}_{1} , {y}_{1}\right) = \left(1 , 3\right) , \left({x}_{2} , {y}_{2}\right) = \left(- 2 , - 5\right)$

$d = \sqrt{{\left(- 2 - 1\right)}^{2} + {\left(5 - 3\right)}^{2}} = \sqrt{9 + 64} = \sqrt{73} \approx 8.544$

$\textcolor{b l u e}{\text{sum of radii }} = 1 + 2 = 3$

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

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

$\Rightarrow \text{smallest distance } = 8.544 - 3 = 5.544$
graph{(y^2-6y+x^2-2x+9)(y^2+10y+x^2+4x+25)=0 [-12.49, 12.48, -6.24, 6.25]}