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  • Two rods of lengths a and b slide along the axes which are rectangular in such a manner that their ends are concyclic. Then the locus of the centre of circle passing through these ends is

Two rods of lengths a and b slide along the axes which are rectangular in such a manner that their ends are concyclic. Then the locus of the centre of circle passing through these ends is the curve.

Option: 1

\mathrm{4\left(x^2-y^2\right)=a^2-b^2}


Option: 2

\mathrm{2\left(x^2-y^2\right)=a^2+b^2}


Option: 3

\mathrm{4\left(x^2-y^2\right)=a^2+b^2}


Option: 4

\mathrm{x^2-y^2=a^2-b^2}


Answers (1)

best_answer

Let the two rods be AB(along x-axis) and CD along y-axis so that four points A, B, C, D are on a circle whose centre is (h, k), say. If r be radius then from the figure we have.

\mathrm{ r^2=P A^2=k^2+a^2 / 4 . }
\mathrm{ r^2=P C^2=h^2+\frac{b^2}{4} . }
Eliminating r by subtracting, we get
\mathrm{ h^2-k^2=\frac{a^2}{4}-\frac{b^2}{4} }
or \mathrm{ 4\left(x^2-y^2\right)=a^2-b^2 } is the required locus.

Posted by

vishal kumar

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