Find the equation of a circle with

The general form of the equation of a circle is:

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Substituting the centre and radius of the circle in he general form:

$(x-(a \cos \propto))^{2}+(y-(a \sin \alpha))^{2}=a^{2}$

$\Rightarrow(x-a \cos \alpha)^{2}+(y-a \sin \alpha)^{2}=a^{2}$

$\Rightarrow x^{2}-2 x a \cos \alpha+a^{2} \cos ^{2} \alpha+y^{2}-2 y a \sin \alpha+a^{2} \sin ^{2} \alpha=a^{2}$

$\Rightarrow x^{2}+y^{2}+a^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)-2 a(x \cos \alpha+y \sin \alpha)=a^{2}$

$\Rightarrow x^{2}+y^{2}+a^{2}-2 a(x \cos \alpha+y \sin \alpha)=a^{2} \ldots\left(\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)=1\right)$

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

Ans: equation of a circle with Centre $(a \cos \propto, a \sin \propto)$ and radius a is:

$x^{2}+y^{2}-2 a(x \cos a+y \sin a)=0$