Prove that the locus of the center of a circle , which intercepts a cord of given length 2a on the axis of x and passes through a given point on the axis of y distant b from the origin , is the curve x^2-2yb+b^2=a^2

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Arun · Answer

We know length of intercept made by circle on X axis is 2√( G 2 -C )= 2A AND. Length of intercept made by circle on y axis is 2 √(F 2 --c)=b Just do the squaring and add them you will get the same equation

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Prove that the locus of the center of a circle , which intercepts a cord of given length 2a on the axis of x and passes through a given point on the axis of y distant b from the origin , is the curve x^2-2yb+b^2=a^2

Prove that the locus of the center of a circle , which intercepts a cord of given length 2a on the axis of x and passes through a given point on the axis of y distant b from the origin , is the curve x^2-2yb+b^2=a^2

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Prove that the locus of the center of a circle , which intercepts a cord of given length 2a on the axis of x and passes through a given point on the axis of y distant b from the origin , is the curve x^2-2yb+b^2=a^2

Prove that the locus of the center of a circle , which intercepts a cord of given length 2a on the axis of x and passes through a given point on the axis of y distant b from the origin , is the curve x^2-2yb+b^2=a^2

1 Answers

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