Question

Show that the equation to the tangent at $A$ where $\theta =\alpha$ to the circle $(x-a{)}^2+(y-b{)}^2=r^2$ is $(x-a)\cos \alpha +(y-b)\sin \alpha =r$.

Answer 1

Shift the origin to the point $(a,b)$ by writing $x+a$ of $x$ and $y+b$ for $y$ so that the equation of the circle becomes
$x^2+y^2=r^2$.
Tangent at $\theta =\alpha$ i.e. $(r\cos \alpha ,r\sin \alpha )$ is
$x\left(r\cos \alpha \right)+y\left(r\sin \alpha \right)=r^2$
or $x\cos \alpha +y\sin \alpha =r$.
Shift the origin back to $(a,b)$ by writing $x-a$ for $x$ and $y-b$ for $y$.
$\therefore (x-a)\cos \alpha +(y-b)\sin \alpha =r$
is the required tangent.