Question

The equation of the circle and its chord are respectively $x^2+y^2=a^2$ are $xcos\alpha +y\sin \alpha =p.$ The equation of the circle of which this chord is diameter is

• A. $x^2+y^2-2px\cos \alpha -2py\sin \alpha +2p^2-a^2=0$
• B. $x^2+y^2-2px\cos \alpha -2py\sin \alpha +p^2-a^2=0$
• C. $x^2+y^2-2px\cos \alpha +2py\sin \alpha +2p^2-a^2=0$
• D. None of these

Answer 1

The correct option is A $x^2+y^2-2px\cos \alpha -2py\sin \alpha +2p^2-a^2=0$

The equation of any circle through their points of intersection is

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

Its centre $=(\frac{-\lambda \cos \alpha }{2},\frac{-\lambda \sin \alpha }{2})$

But the line $x\cos \alpha +y\sin \alpha =p$ is a diameter of the circle.

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

$\Rightarrow \lambda =-2p$

Thus, The equation of the required circle is

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