Question

If the locus of the centre of a circle which touches the line $x\cos \alpha +y\sin \alpha =p$ and the circle $(x-a{)}^2+(y-b{)}^2=c^2$ is $(x-a{)}^2+(y-b{)}^2=(x\cos \alpha +y\sin \alpha +k{)}^2$ then $k=$

• A. $p$
• B. $-p\pm c$
• C. $pc$
• D. $-p$

Answer 1

Answer: (B)

$-p\pm c$

If $(x-a{)}^2+(y-b{)}^2=(x\cos \alpha +y\sin \alpha +k{)}^2$ touches $(x-a{)}^2+(y-b{)}^2=c^2$ then $c^2=(x\cos \alpha +y\sin \alpha +k{)}^2$
$x\cos \alpha +y\sin \alpha +k=c$ or $x\cos \alpha +y\sin \alpha +k=-c$
$p+k=c$ or $p+k=-c$
$k=c-p$ or $k=-c-p$
So, $k=-p{+}_{-}c$