Q.60. Circles with radii increasing with time. Time of touching each other.

Question 60.
If the radius of the circle (x-1)^2 + (y-2)^2 = 1 and (x-7)^2 + (y-10)^2 = 4 are increasing uniformly w.r.t time as 0.3 and 0.4 unit/sec, then at which time t will they touch each other ?

Answer 60.
Let them touch each other at time = t s.
Radius of the first circle will be 1 + 0.3t and that of the second will be 2 + 0.4t

Distance between the centres of the two circles at t = 0 s
= √[(1 - 7)^2 + (2 - 10)^2] = 10
This is more than the distance between the radii of the two circles = 1 + 2 = 3
=> at t s, the circles will touch externally and at that time
sum of the radii = 10
=> 1 + 0.3t + 2 + 0.4t = 10
=> 0.7t = 7
=> t = 7/0.7 = 10 s.

Again they will alslo touch each other internally when the distance between the centres equals the difference of the radii
=> 10 = (2 + 0.4t) - (1 + 0.3t) = 1 + 0.1t
=> 0.1t = 9
=> t = 90 s.

Answer:
At t = 10 s, they will touch each other externally
and at t = 90 s, they will touch each other internally.

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