Question

lf the line xcos $\alpha +$ ysin $\alpha =p$ and the circle $x^2+y^2=a^2$ intersect at $A$ and $B$ then the equation of the circle on $AB$ as diameter $(x^2+y^2-a^2)+k(x\cos \alpha +y\sin \alpha -p)=0$ then $k=$

• A. $p$
• B. $-p$
• C. $-4p$
• D. $-2p$

Answer 1

Answer: (D)

$-2p$

Centre of the circle having $AB$ as diameter is the foot of perpendicular from $(0,0)$ to $xcos\alpha +ysin\alpha =p$ then
$\frac{x-0}{cos\alpha }=\frac{y-0}{sin\alpha }=-\frac{(-p)}{1}$
$x=pcos\alpha$ and $y=psin\alpha$
Equation of circle having centre $(pcos\alpha ,psin\alpha )$ and
$radius\left(r\right)=\sqrt{a^2-\left(\sqrt{p^2(si{n}^2\alpha +co{s}^2\alpha )}\right)2}$
$=\sqrt{a^2-p^2}$
$\therefore (x-pcos\alpha {)}^2+(y-psin\alpha {)}^2=(\sqrt{a^2-p^2}{)}^2$
$x^2+y^2-a^2+p^2-x2pcos\alpha -2pysin\alpha +p^2=0$
$(x^2+y^2-a^2)-2p(2cos\alpha x+2sin\alpha y-p)=0$
So, $k=-2p$
$\therefore AB=2r$