Question

The locus of the centre of the circle $xcos\alpha +ysin\alpha =a$ and $xsin\alpha +ycos\alpha =b$

• A. $x^2+y^2=a^2+b^2$
• B. $x^2+y^2=a^2{b}^2$
• C. $x^2-y^2=a^2+b^2$
• D. $x^2+y^2=a^2-b^2$

Answer 1

Answer: (A)

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

We have,

Given two equations are

$x\cos \alpha +y\sin \alpha =a.......\left(1\right)$

$x\sin \alpha -y\cos \alpha =b.......\left(2\right)$

Squaring both side and adding we get,

${(x\cos \alpha +y\sin \alpha )}^2+{(x\sin \alpha -y\cos \alpha )}^2=a^2+b^2$

$x^2{\cos }^2\alpha +y^2{\sin }^2\alpha +2xy\sin \alpha \cos \alpha +x^2{\sin }^2\alpha +y^2{\cos }^2\alpha -2xy\sin \alpha \cos \alpha =a^2+b^2$

$x^2{\cos }^2\alpha +y^2{\sin }^2\alpha +x^2{\sin }^2\alpha +y^2{\cos }^2\alpha =a^2+b^2$

$x^2({\cos }^2\alpha +{\sin }^2\alpha )+y^2({\cos }^2\alpha +{\sin }^2\alpha )=a^2+b^2$

$x^2\times 1+y^2=a^2+b^2$

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

Hence, this is the locus of circle.

This is the answer.