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What's the centre of circles

+3 votes

which always passes through the fixed points (a,0) and (-a,0) ?

asked Feb 8, 2013 in PRECALCULUS by andrew Scholar

1 Answer

+3 votes

Passes through the fixed points are (a, 0) and (-a, 0)

Mid point of center of circle is [(a-a)/2, (0+0)/2] = [0, 0]

The radius is distance between the two numbers.

So, r = √[(a-(-a))2+ (0-0)2]

r = √[2a]2  = 2a

Cartesian coordinates

In an xy Cartesian coordinate system, e circle with Centre coordinates (a, b) and radius r is the set of all points (x, y) such that  (x - a)2 + (y - b)2 = r2

point (0, 0) and radius r = 2a.

So, (x - 0)2 + (y - 0)2 = (2a)2

x2 + y2 = 4a2.

answered Feb 8, 2013 by richardson Scholar

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