The shortest distance between the circles x^2 + y^2 =144 and (x-3)^2 + (y-4)^2 =25 is- Share with your friends Share 1 Manvendra S. answered this Dear Student, Please find below the solution to the asked query: Circle1 is x2+y2=144 have center 0,0 and radius is 12Circle2 is x-32+y-42=25 have center 3,4 and radius is 5So in the figure OA =10 unit and OB=12 unit Hence the minimum distance will be OB-OA⇒12-10⇒2unit AnswerHope this information will clear your doubts about the topic. If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible. Regards -1 View Full Answer Burnet answered this The circle lies inside the circle x^2 + y^2=144 with radius 12 Shortest distance will be 12-the diameter of the circle That is 12-10=2 1
