Circle A has a radius of #6 # and a center of #(4 ,3 )#. Circle B has a radius of #3 # and a center of #(1 ,8 )#. If circle B is translated by #<-2 ,4 >#, does it overlap circle A? If not, what is the minimum distance between points on both circles?
1 Answer
no overlap, min distance ≈1.296
Explanation:
What we have to do here is
#color(blue)"compare"# the distance (d ) between the centres of the circles to the#color(blue)"sum of the radii"# • If sum of radii > d , then circles overlap
• If sum of radii < d , then no overlap
Before calculating d we require to find the 'new' centre of B under the given translation which does not change the shape of the circle only it's position.
#"Under a translation "((-2),(4))#
#(1,8)to(1-2,8+4)to(-1,12)larrcolor(blue)" new centre of B"# To calculate d, use the
#color(blue)"sistance formula"#
#color(red)(bar(ul(|color(white)(2/2)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(2/2)|)))#
where# (x_1,y_1),(x_2,y_2)" are 2 coordinate points"# The 2 points here are (4 ,3) and (-1 ,12)
let
# (x_1,y_1)=(4,3)" and " (x_2,y_2)=(-1,12)#
#d=sqrt((-1-4)^2+(12-3)^2)=sqrt(25+81)≈10.296# Sum of radii = radius of A + radius of B = 6 + 3 = 9
Since sum of radii < d , then circles do not overlap
#"min. distance between points "=d-" sum of radii"#
#rArr"min. distance "=10.296-9=1.296#
graph{((x-4)^2+(y-3)^2-36)((x+1)^2+(y-12)^2-9)=0 [-56.2, 56.2, -28.1, 28.1]}
