Circle A has a radius of $3$ $3$ and a center of $(3, 9)$ $(3 ,9 )$ . Circle B has a radius of $2$ $2$ and a center of $(1, 4)$ $(1 ,4 )$ . If circle B is translated by $< 3, - 1 >$ $<3 ,-1 >$ , does it overlap circle A? If not, what is the minimum distance between points on both circles?

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Answer 1 · 2016-09-17T03:23:44

No overlap between Circle A and Circle B. Minimum distance = 1.083. See explanation and the diagram below.

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Compare the distance (d) between the centres of the circles to the sum of the radii.

1) If the sum of the radii $>$ $>$ d, the circles overlap.
2) If the sum of the radii $<$ $<$ d, the no overlap.

The first step here is to calculate the new centre of B under the traslation, which does not change the shape of the circle only its' position.

Under a translation $< 3, - 1 >$ $<3, -1>$
$B (1, 4) \Rightarrow (1 + 3, 4 - 1) \Rightarrow (4, 3)$ $B(1,4) => (1+3,4-1) => (4,3)$ new centre of B

To calculate d, use the distance formula :
$d = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

where $(x_{1}, y_{1}) and (x_{2}, y_{2})$ $(x_1,y_1) and (x_2,y_2)$ are 2 coordinate points

here the two points are $(3, 9)$ $(3,9)$ and $(4, 3)$ $(4,3)$ the centres of the circles

let $(x_{1}, y_{1}) = (3, 9)$ $(x_1,y_1)=(3,9)$ and $(x_{2}, y_{2}) = (4, 3)$ $(x_2,y_2)=(4,3)$

$d = \sqrt{{(4 - 3)}^{2} + {(3 - 9)}^{2}} = \sqrt{1^{2} + {(- 6)}^{2}} = \sqrt{37} = 6.083$ $d=sqrt((4-3)^2+(3-9)^2)=sqrt(1^2+(-6)^2)=sqrt37=6.083$

Sum of radii = radius of A + radius of B $= 3 + 2 = 5$ $= 3+2=5$

Since sum of radius $<$ $<$ d, then no overlap of the circles

Min.D minimum distance (the red line in the daigram) :

$d -$ $d-$ sum of radii $= 6.083 - 5 = 1.083$ $= 6.083-5=1.083$

Circle A has a radius of 3 33 and a center of (3 ,9 )(3,9)(3 ,9 ). Circle B has a radius of 2 22 and a center of (1 ,4 )(1,4)(1 ,4 ). If circle B is translated by <3 ,-1 ><3,−1><3 ,-1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?