Given are two circles whose radii are given, such that the smaller lies completely within the bigger circle, and they touch each other at one point. We have to find the length of the perpendicular bisector of the line joining the centres of the circles.
Input: r1 = 5, r2 = 3 Output: 9.79796 Input: r1 = 8, r2 = 4 Output: 15.4919
- Let the two circles have center at A and B.The perpendicular bisector PQ, bisects the line at C.
- Let radius of bigger circle = r1
radius of smaller circle = r2
- so, AB = r1-r2,
- therefore, AC = (r1-r2)/2
- In, the figure, we see
PA = r1
- in triangle ACP,
PC^2 + AC^2 = PA^2
PC^2 = PA^2 – AC^2
PC^2 = r1^2 – (r1-r2)^2/4
- so, PQ = 2*√(r1^2 – (r1-r2)^2/4)
Length of the perpendicular bisector = 2 * sqrt(r1^2 – (r1-r2)*(r1-r2)/4)
Below is the implementation of the above approach:
The length of the perpendicular bisector is 9.79796
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