Question

Two rods of lengths $a$ and $b$ slide along coordinate axes such that their ends are concyclic. Locus of the center of the circle is

• A. $4(x^2+y^2)=a^2+b^2$
• B. $4(x^2+y^2)=a^2-b^2$
• C. $4(x^2-y^2)=a^2-b^2$
• D. $xy=ab$

Answer 1

The correct option is C $4(x^2-y^2)=a^2-b^2$

For the rod of length b

the extremes point (o, q) and

(o, q + b) and for the rod of

length a the extreme points

(p,o) and (p, p + a). The center of

circle will be intersection between

the $\perp$ eq bisectors of rods

with length a and b.

$(x,y)=(p+\frac{a}{2},q+\frac{b}{2})$

So

$x=p+\frac{a}{2}$ $y=q+\frac{b}{2}$

$x-p=\frac{a}{2}$ $y-q=\frac{b}{2}$ __(1)

Now we need radius

But distance between (x, y) and

(p,o) and also distance between

(x,y) and (o,q) so

$(x-p{)}^2+y^2=x^2+(y-q{)}^2$ __(2)

Now

$(\frac{a}{2}{)}^2+y^2=x^2+(\frac{b}{2}{)}^2$

$\frac{a^2-b^2}{4}=x^2-y^2$

$a^2-b^2=4(x^2-y^2)$