Circle A has a radius of $2$ $2$ and a center of $(6, 5)$ $(6 ,5 )$ . Circle B has a radius of $3$ $3$ and a center of $(2, 4)$ $(2 ,4 )$ . If circle B is translated by $< 1, 1 >$ $<1 ,1 >$ , does it overlap circle A? If not, what is the minimum distance between points on both circles?

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Circle A has a radius of $2$ $2$ and a center of $(6, 5)$ $(6 ,5 )$ . Circle B has a radius of $3$ $3$ and a center of $(2, 4)$ $(2 ,4 )$ . If circle B is translated by $< 1, 1 >$ $<1 ,1 >$ , does it overlap circle A? If not, what is the minimum distance between points on both circles?

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Answer 1 · 2018-05-14T09:25:40

$circles overlap$ $"circles overlap"$

Explanation:

$what we have to do here is compare the distance (d)$ $"what we have to do here is compare the distance (d)"$
$between the centres to the sum of the radii$ $"between the centres to the sum of the radii"$

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

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

$before calculating d we require to find the new centre$ $"before calculating d we require to find the new centre"$
$of B after the given translation$ $"of B after the given translation"$

$under the translation < 1, 1 >$ $"under the translation "< 1,1>$

$(2, 4) \to (2 + 1, 4 + 1) \to (3, 5) \leftarrow new centre of B$ $(2,4)to(2+1,4+1)to(3,5)larrcolor(red)"new centre of B"$

$to calculate d use the distance formula$ $"to calculate d use the "color(blue)"distance formula"$

$d = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$let (x_{1}, y_{1}) = (6, 5) and (x_{2}, y_{2}) = (3, 5)$ $"let "(x_1,y_1)=(6,5)" and "(x_2,y_2)=(3,5)$

$d = \sqrt{{(3 - 6)}^{2} + {(5 - 5)}^{2}} = \sqrt{9} = 3$ $d=sqrt((3-6)^2+(5-5)^2)=sqrt9=3$

$sum of radii = 2 + 3 = 5$ $"sum of radii "=2+3=5$

$since sum of radii > d then circles overlap$ $"since sum of radii">d" then circles overlap"$
graph{((x-6)^2+(y-5)^2-4)((x-3)^2+(y-5)^2-9)=0 [-20, 20, -10, 10]}

Answer 2 · 2018-05-14T11:08:36

The distance between the centers is $3$ $3$ , which satisfies the triangle inequality with the two radii of $2$ $2$ and $3$ $3$ , so we have overlapping circles.

Explanation:

I thought I did this one already.

A is $(6, 5)$ $(6,5)$ radius $2$ $2$

B's new center is $(2, 4) + < 1, 1 > = (3, 5),$ $(2,4)+<1,1> =(3,5),$ radius still $3$ $3$

Distance between centers,

$d = \sqrt{{(6 - 3)}^{2} + {(5 - 5)}^{2}} = 3$ $d = sqrt{ (6-3)^2 + (5-5)^2}=3$

Since the distance between the centers is less than the sum of the two radii, we have overlapping circles.

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Circle A has a radius of $2$ $2$ and a center of $(6, 5)$ $(6 ,5 )$ . Circle B has a radius of $3$ $3$ and a center of $(2, 4)$ $(2 ,4 )$ . If circle B is translated by $< 1, 1 >$ $<1 ,1 >$ , does it overlap circle A? If not, what is the minimum distance between points on both circles?

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Explanation:

Explanation:

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Circle A has a radius of 22 and a center of (6,5)(6 ,5 ). Circle B has a radius of 33 and a center of (2,4)(2 ,4 ). If circle B is translated by <1,1><1 ,1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

2 Answers

Explanation:

Explanation:

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Circle A has a radius of $2$ $2$ and a center of $(6, 5)$ $(6 ,5 )$ . Circle B has a radius of $3$ $3$ and a center of $(2, 4)$ $(2 ,4 )$ . If circle B is translated by $< 1, 1 >$ $<1 ,1 >$ , does it overlap circle A? If not, what is the minimum distance between points on both circles?