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Two equal circles of radius r intersect such that each passes through the centre of the other. The length of common chord of the circles is

Question

Two equal circles of radius r intersect such that each passes through the centre of the other.

The length of the common chord of the circles is:

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Solution

The two circles intersect at two points creating an isosceles triangle with the line connecting the centers of the circles as the base. The common chord is the base of the two equal right-angled triangles.

Step 1: Draw the two intersecting circles with a common chord.

Step 2: Draw a line from the center of one circle to the center of the other. This line is 2r (twice the radius).

Step 3: Draw lines from the centers of the circles to the endpoints of the common chord. You now have two right-angled triangles.

Step 4: The hypotenuse of each triangle is the radius of the circles (r). The base of the triangle is half the length of the common chord. The other side of the triangle is the distance between the center of a circle and the common chord, which is also r.

Step 5: Use the Pythagorean theorem to solve for the length of the common chord. The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

So, r^2 = r^2 + (common chord/2)^2

Solving for the common chord gives:

common chord = 2 * sqrt(2*r^2 - r^2) = 2 * r * sqrt(2 - 1) = r * sqrt(2)

So, the length of the common chord of the circles is r * sqrt(2).

This problem has been solved

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