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The locus of the center of the circle which bisect the circumference of the circles x^2 y^2=4 & x^2 y^2-2x 6y 1=0?
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The locus of the center of the circle which bisect the circumference o...
The equation of the circle with center (h,k) and radius r is given by (x-h)^2 + (y-k)^2 = r^2.

In this case, we want to find the locus of the centers of the circles that bisect the circumference of the circle x^2 + y^2 = 4.

Let (a, b) be the coordinates of the center of the circle we are looking for. Since the circle bisects the circumference of the circle x^2 + y^2 = 4, the distance from (a, b) to the origin (0, 0) is equal to the radius of the larger circle, which is 2.

Using the distance formula, we have:

√((a-0)^2 + (b-0)^2) = 2

Simplifying this equation, we get:

√(a^2 + b^2) = 2

Squaring both sides, we have:

a^2 + b^2 = 4

This is the equation of a circle centered at the origin (0, 0) with radius 2. Therefore, the locus of the centers of the circles that bisect the circumference of x^2 + y^2 = 4 is the circle x^2 + y^2 = 4.
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The locus of the center of the circle which bisect the circumference of the circles x^2 y^2=4 & x^2 y^2-2x 6y 1=0?
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