In the xy-plane, the circles with equations (x-3)^2 + y^2 = 11 and (x+2)^2 + y^2 = r^2 are externally tangent. What is the value of r?

For the circle with equation (x - 3)² + y² = 11,
Center = (3, 0) and radius = √11

For the circle with equation (x + 2)² + y² = r²,
Center = (-2, 0) and radius = r

The circles are externally tangent, and thus
Distance between the two center = Sum the two radii
3 - (-2) = √11 + r
r = 5 - √11
r = 1.7

The answer: C) 1.7

C1: (x-3)^2 + y^2 =  11...(1), center of C1 is ( 3,0), r1= rt11.;
C2: (x+2)^2 +y^2 = r^2...(2), center of C2 is (-2,0), r2= r.;
Since circles C1 & C2 are externally tangential, The distance between their 
centers = the sum of their radii, ie., 3-(-2) = 5 = r + rt11. Then r = 5-rt11 = 1.68337521 = 1.7 rounded to 1 decimal place.

Find the distance between the centers of the circles

centers are (3 , 0) and (-2 , 0)

The distance between the centers is 5.

The distance between the centers is equal to the sum of the radii of the circles

5 = sqrt(11) + r
5 - sqrt(11) = r
r = 1.683375209644600150885067263329....

Pick the most appropriate choice.