Two straight rods of lengths \(2 a\) and \(2 b\) move along the coordinate axes in such a way that their…

Two straight rods of lengths \(2 a\) and \(2 b\) move along the coordinate axes in such a way that their extremities are always concyclic. Then the locus of the centres of such circles is
  1. \(2\left(x^2+y^2\right)=a^2+b^2\)
  2. \(2\left(x^2-y^2\right)=a^2+b^2\)
  3. \(x^2+y^2=a^2+b^2\)
  4. \(x^2-y^2=a^2-b^2\)

Solution

According to given information, if we draw the figure.
Let the equation of circle is \(\begin{aligned} & x^2+y^2+2 g x+2 f y+c=0 \\ & \because 2 \sqrt{g^2-c}=2 a \\ & \text { and } 2 \sqrt{f^2-c}=2 b \\ & \text { then } g^2-a^2=0 \text { and } f^2-b^2=0 \\ & \text { so, } g^2-a^2=f^2-b^2 \\ & \Rightarrow g^2-f^2=a^2-b^2 \end{aligned}\) On taking locus of the centre \((-g,-f)\), we get \(x^2-y^2=a^2-b^2\) Hence, option (4) is correct.

Asked in: AP EAMCET 2019 (20 Apr Shift 1)

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