Question

The radius of the circle ${(x\cos \theta +y\sin \theta -a)}^2+(x\sin \theta -y\cos \theta )=k^2$ is

• A. $a^2+b^2-k^2$
• B. $asin\theta -b\cos \theta$
• C. $a^2+b^2$
• D. $k$

Answer 1

Answer: (D)

Given equation of circle is ${(x\cos \theta +y\sin \theta -a)}^2+{(x\sin \theta -y\cos \theta -b)}^2$
$=k^2$
$\Rightarrow$ $x^2{\cos }^2\theta +y^2{\sin }^2\theta +2xy\sin \theta .\cos \theta +a^2$
$-2a(x\cos \theta +y\sin \theta )+x^2{\sin }^2\theta +y^2{\cos }^2\theta$
$-2xy\sin \theta .\cos \theta +b^2-2b(x\sin \theta -ycos\theta )=k^2$
$\Rightarrow$ $x^2({\sin }^2\theta +{\cos }^2\theta )+y^2({\sin }^2\theta +{\cos }^2\theta )$
$+(-2a\cos \theta -2b\sin \theta )x+(-2a\sin \theta +2b\cos \theta )y$
$+(a^2+b^2-k^2)=0$
$\Rightarrow$ $x^2+y^2-2(a\cos \theta +b\sin \theta )x$ $+2(-a\sin \theta +b\cos \theta )y+(a^2+b^2-k^2)=0$ ..(i)
$(\because {\sin }^2\theta +{\cos }^2\theta =1)$
On comparing with
$x^2+y^2+2gx+2fy+c=0,$ we get
$g=-(a\cos \theta +b\sin \theta ),$
$f=+(-a\sin \theta +b\cos \theta )$
and $c=a^2+b^2-k^2$
$\therefore$ Radius $=\sqrt{g^2+f^2-c}$
$=\sqrt{\begin{array}{rl}& {(a\cos \theta +b\sin \theta )}^2+{(-a\sin \theta +b\cos \theta )}^2\\ & -(a^2+b^2-k^2)\end{array}}$
$=\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\sin \theta .\cos \theta }$
$+a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\sin \theta .\cos \theta$
$-(a^2+b^2-k^2)$
$\Rightarrow$ $\sqrt{a^2+b^2-a^2-b^2+k^2}=\sqrt{k^2}=k$