Assume that the origin (0,0) is the centre of the smallest circle and that the centre of the semicircle is (d,0); hence its radius r = d+1 Now it can be easily verified that the semicrcle and the smallest circle x^2+y^2=1 touch each other at (-1,0) as depicted. Our assumption that the centre of the semicircle lies on the x-axis seems, therefore, to be correct. Now let the semicircle intesect x^2+y^2 =4 in (x1,y1) in the first or second quadrant and let it intersect x^2+y^2=9 in the fourth quadrant in (x2,y2). We can easily show that x1=(4-d)/d while x2=(3-2d)/2d. Further, (x1+x2)=2d since the midpoint of (x1,y1) and (x2,y2) is (d,0). Thus,(4-d)/d + (3-2d)/2d =2d which gives d = V3 -1/2 or r = d+1 = V3 + 1/2 (Option D) Would that be correct? Ajit
Both answers are similar and correct. We actually had a little shorter solution like: cos(a)=-cos(180-a) which immediately gives a quadratic eqn leading to the solution, but we've lost it. And for some reason I just can't see it!!! Anyways... Thanks you both for the solution.
Assume that the origin (0,0) is the centre of the smallest circle and that the centre of the semicircle is (d,0); hence its radius r = d+1
ReplyDeleteNow it can be easily verified that the semicrcle and the smallest circle x^2+y^2=1 touch each other at (-1,0) as depicted. Our assumption that the centre of the semicircle lies on the x-axis seems, therefore, to be correct. Now let the semicircle intesect x^2+y^2 =4 in (x1,y1) in the first or second quadrant and let it intersect x^2+y^2=9 in the fourth quadrant in (x2,y2). We can easily show that x1=(4-d)/d while x2=(3-2d)/2d. Further, (x1+x2)=2d since the midpoint of (x1,y1) and (x2,y2) is (d,0). Thus,(4-d)/d + (3-2d)/2d =2d which gives d = V3 -1/2 or r = d+1 = V3 + 1/2 (Option D)
Would that be correct?
Ajit
Erratum: That's Option E
ReplyDeleteAjit
Looks good to me.
ReplyDeletehttp://i38.tinypic.com/zvu244.jpg
ReplyDeleteBoth answers are similar and correct.
ReplyDeleteWe actually had a little shorter solution like:
cos(a)=-cos(180-a)
which immediately gives a quadratic eqn leading to the solution, but we've lost it. And for some reason I just can't see it!!!
Anyways...
Thanks you both for the solution.