Circle cuts circle - (3)
Two circles of radii one and two units respectively are orthogonal to each other. If the length of the common chord of these two circles can be expressed in the form where is square free , find .
This question is a part of this set (click here) .Follow me for more! :)
The answer is 9.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
The two equations for the two circles are as follows:
Circle with radius = 1 -> x^2 + y^2 = 1
Circle with radius = 2 -> (x-sqrt(5))^2 + y^2 = 4 (This circle is orthogonal to the first circle)
When we say Circle A is orthogonal to circle B, Then it means that the tangents of these circles at the points of intersection are perpendicular to each other.
Solving for x and y which satisfies both these equations, we find that:
The common chord for these two circle passes through the points (0.447,0.894) and (0.447,-0.894)
The length of this common chord is 0.894 - (-0.894) = 1.788 = 4*sqrt(5)/5
So a + b = 4 + 5 = 9