Two thin rods AB and CD of length 2a and 2b moves along OX and OY where O is the origin. Find the locus of the center of the circle passing through the extremities of the two rods.
My attempt:-
There are two points (2a,0) and (0,2b) through which the circle passes,
So let the center C be (h,k) and radius r
$(h-2a)^2 + k^2 = r^2$
$h^2 + (k-2b)^2 = r^2$
$(h-2a)^2 + k^2 = h^2 + (k-2b)^2$
$-4ax+4a^2 = -4bk + 4b^2$
$a^2-b^2 = ax-bk$
I don't know what to do next.
The correct answer is $x^2-y^2 = a^2-b^2$
Best Answer
My attempt
Length of intercept on y-axis
$$ 2b =2√(g^2-c)$$
Length of intercept on x-axis
$$ 2a =2√(f^2-c)$$ From the above equations you get
$$(g^2 - a^2)=(f^2-b^2)$$ From this we would get
$$g^2 -f^2 = a^2 -b^2$$
where (g,f) is the centre of the circle. Replace (g,f) by (x,y), and you would get the answer:
$$ x^2 -y^2 = a^2 - b^2 $$