Circle A has a center at #(-8 ,8 )# and a radius of #8 #. Circle B has a center at #(-3 ,3 )# and a radius of #4 #. Do the circles overlap? If not, what is the smallest distance between them?

Question

Circle A has a center at #(-8  ,8   )# and a radius of #8    #.  Circle B has a center at #(-3  ,3   )# and a radius of #4 #.  Do the circles overlap?  If not, what is the smallest distance between them?

Allison Allen · Accepted Answer

the circles will overlap

The radius of circle A is #r_A=8# unit The radius of circle B is #r_B=4# unit The distance between their centers #d_(AB)=sqrt((-8+3)^2+(8-3)^2)=5sqrt2#unit It is obvious that #r_A+r_B>d_(AB)# Hence the circles will overlap Further If #r_A+r_B=d_(AB)# then they would touch AND #r_A+r_B< d_(AB)# then they would not touch or overlap In that case the would have minimum distance #d_(min)=d_(AB)-(r_A+r_B)# Is it OK?

Evelyn Arboleda · Answer

undefined To determine if the circles overlap, you can calculate the distance between their centers and compare it to the sum of their radii. If the distance between the centers is greater than the sum of the radii, the circles do not overlap.

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

For Circle A with center $(-8, 8)$ and Circle B with center $(-3, 3)$, the distance between their centers is:

$$d = \sqrt{(-3 - (-8))^2 + (3 - 8)^2}$$
$$d = \sqrt{(5)^2 + (-5)^2}$$
$$d = \sqrt{25 + 25}$$
$$d = \sqrt{50}$$
$$d = 5\sqrt{2}$$

Now, compare this distance to the sum of the radii of the circles. Circle A has a radius of 8 and Circle B has a radius of 4. So, the sum of their radii is $8 + 4 = 12$.

Since $5\sqrt{2} > 12$, the circles do not overlap.

To find the smallest distance between them, subtract the sum of their radii from the distance between their centers:

$$5\sqrt{2} - 12$$

Therefore, the smallest distance between the circles is $5\sqrt{2} - 12$.