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વર્તૂળ અને તેની જીવાનું સમીકરણ અનુક્રમે $x^2 + y^2 = a^2$ અને $x\ cos\ \alpha + y\ sin\ \alpha = p$ છે. આ જીવા જે વર્તૂળનો વ્યાસ હોય તે વર્તૂળનું સમીકરણ :
$x^2 + y^2 - 2px\ cos\ \alpha - 2py\ sin\ \alpha + 2p^2- a^2 =0$
$x^2 + y^2- 2px\ cos\ \alpha - 2py\ sin\ \alpha + p^2- a^2 = 0$
$x^2 + y^2 - 2px\ cos\ \alpha + 2py\ sin\ \alpha + 2p^2 - a^2 =0$
એકપણ નહિ
Solution
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 $=\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)$
But the line $x \cos \alpha+y \sin \alpha=p$ is a diameter of the circle.
$\therefore\left(\frac{-\lambda \cdot \cos \alpha}{2}\right) \cos \alpha+\left(\frac{-\lambda \sin \alpha}{2}\right) \sin \alpha-p=0$
$\Rightarrow \lambda=-2 p$
Thus, The equation of the required circle is
$x^2+y^2-a^2-2 p(x \cos \alpha+y \sin \alpha-p)=0$
