Two straight rods of lengths 2a and 2b 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

Question

Two straight rods of lengths 2a and 2b 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 (A) 2 (x2 + y2) = a2 + b2 (B) 2 (x2 - y2) = a2 + b2 (C)  x2 + y2 = a2 + b2 (D)  x2 - y2 = a2 - b2

Tardigrade Admin · Accepted Answer

According to given information, if we draw the figure.

Let the equation of circle is

x^{2} + y^{2} + 2 gx + 2 f y + c = 0

∵ 2 g^{2} - c = 2 a

and

2 f^{2} - c = 2 b

then

g^{2} - a^{2} = 0

and

f^{2} - b^{2} = 0

so,

g^{2} - a^{2} = f^{2} - b^{2}

\Rightarrow g^{2} - f^{2} = a^{2} - b^{2}

On taking locus of the centre

(- g, - f)

, we get

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

Q. 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

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

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

$x^{2} + y^{2} = a^{2} + b^{2}$

$x^{2} - y^{2} = a^{2} - b^{2}$

Q. Two straight rods of lengths 2a and 2b 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

2(x2+y2)=a2+b2

2(x2−y2)=a2+b2

x2+y2=a2+b2

x2−y2=a2−b2

According to given information, if we draw the figure. Let the equation of circle is x2+y2+2gx+2fy+c=0 ∵2g2−c​=2a and 2f2−c​=2b then g2−a2=0 and f2−b2=0 so, g2−a2=f2−b2 ⇒g2−f2=a2−b2 On taking locus of the centre (−g,−f), we get x2−y2=a2−b2

Q. 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

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

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

$x^{2} + y^{2} = a^{2} + b^{2}$

$x^{2} - y^{2} = a^{2} - b^{2}$