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Last updated date: 06th Dec 2023
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MVSAT Dec 2023

Two rods of length a and b slide along the x-axis and y-axis respectively in such a manner that their ends are concyclic. The locus of the centre of the circle passing through the end point is
A) $4({{x}^{2}}+{{y}^{2}})={{a}^{2}}+{{b}^{2}}$
B) ${{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}$
C) $4({{x}^{2}}-{{y}^{2}})={{a}^{2}}-{{b}^{2}}$
D) ${{x}^{2}}-{{y}^{2}}={{a}^{2}}-{{b}^{2}}$

Answer
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Hint: A set of points are said to be concyclic if they lie on a common circle. All concyclic points are at the same distance from the centre of the circle. Also, a locus is a set of all points, a line, a line segment, a curve whose location satisfies or is determined by one or more specified conditions.

Complete step by step solution:
Let the four points be A, B, C, D. As they are concyclic which means a circle would pass through them.
Let the equation of the circle be
${{x}^{2}}+{{y}^{2}}-2xh-2yk+c=0$ -----(1)
Which means the centre of the circle would be (h,k).
As the rod ‘a’ slide along x-axis which means
$y=0$. So we get
${{x}^{2}}-2xh+c=0$ -------(2)
Let A and B be $({{x}_{1}},0)$ and $({{x}_{2}},0)$. So we can find ${{x}_{1}}$ and ${{x}_{2}}$ by using equation(2).
Now we have to find AB
$AB={{x}_{2}}-{{x}_{1}}=\sqrt{({{x}_{2}}}+{{x}_{1}}{{)}^{2}}-4{{x}_{1}}{{x}_{2}}=a$
                 (Or)
${{({{x}_{2}}+{{x}_{1}})}^{2}}-4{{x}_{1}}{{x}_{2}}={{a}^{2}}$ -------(3)
Now by using equation (2)
${{x}_{1}}+{{x}_{2}}=2h,{{x}_{1}}{{x}_{2}}=c$
Now by putting values of ${{x}_{1}}+{{x}_{2}}$ and ${{x}_{1}}{{x}_{2}}$ in equation (3)
$4{{h}^{2}}-4c={{a}^{2}}$ -----(4)
Similarly rod ‘b’ slides along y-axis which means
$x=0$. So we get
${{y}^{2}}-2yk+c=0$ -----(5)
Let C and D be $(0,{{y}_{1}}) $ and ($(0,{{y}_{2}})$. So we can find ${{y}_{1}}$ and ${{y}_{2}}$ by equation (5)
So, ${{y}_{1}}+{{y}_{2}}=2k,{{y}_{1}}{{y}_{2}}=c$.
Similarly, $C{{D}^{2}}={{({{y}_{2}}-{{y}_{1}})}^{2}}={{({{y}_{1}}+{{y}_{2}})}^{2}}+4{{y}_{1}}{{y}_{2}}={{b}^{2}}$ -(6)
Now by putting the value of ${{y}_{1}}+{{y}_{2}},{{y}_{1}}{{y}_{2}}$ in equation (6)
$4{{k}^{2}}-4c={{b}^{2}}$ ------(7)
Now by subtracting equation (7) from equation (4)
$4{{h}^{2}}-4{{k}^{2}}={{a}^{2}}-{{b}^{2}}$
So the locus of the centre of the circle is option (C) $4({{x}^{2}}-{{y}^{2}})={{a}^{2}}-{{b}^{2}}$.

Note:
Two or three points in the plane that do not all fall on a straight line are concyclic but four or more such points in the plane are not necessarily concyclic. The locus describes the position of points which obey a certain rule.