Question

Equation of a circle drawn in the chord $xcos\alpha +ysin\alpha =p$ of the circle $x^2+y^2=a^2$ as its diameter is

• A. $(x^2+y^2-a^2)-2p(xsin\alpha +ycos\alpha -p)=0$
• B. $(x^2+y^2-a^2)-2p(xcos\alpha +ysin\alpha -p)=0$
• C. $(x^2+y^2-a^2)+2p(xcos\alpha +ycos\alpha -p)=0$
• D. $(x^2+y^2-a^2)-p(xcos\alpha +ysin\alpha -p)=0$

Answer 1

Answer: (A)

$(x^2+y^2-a^2)-2p(xsin\alpha +ycos\alpha -p)=0$

The equation of any circle passing through the points of intersection of $x^2+y^2-a^2=0$

and $x\cos \alpha +y\sin \alpha -p=0$ is

$\Rightarrow x^2+y^2-a^2+\lambda (x\cos \alpha +y\sin \alpha -p)=0$

Where $\lambda$ is a scalar.

Centre of this circle $=(\frac{-\lambda \cos \alpha }{2},\frac{-\lambda \sin \alpha }{2})$

$\because x\cos \alpha +y\sin \alpha -p=0$ is the diameter of the circle, hence the centre must lie on the line.

$\Rightarrow \cos \alpha \left(\frac{-\lambda \cos \alpha }{2}\right)+\sin \alpha \left(\frac{-\lambda \sin \alpha }{2}\right)-p=0$

$\Rightarrow \frac{-\lambda {\cos }^2\alpha }{2}-\lambda {\sin }^2\alpha 2-p=0$

$\Rightarrow p=\frac{-\lambda }{2}({\cos }^2\alpha +{\sin }^2\alpha )$

$\Rightarrow p=\frac{-\lambda }{2}$

or $\lambda =-2p$

$\therefore$ Equation of required circle.

$x^2+y^2-a^2-2p(x\cos \alpha +y\sin \alpha -p)=0$

$\therefore B)$ Answer.