Question

For the circle $(x-h{)}^2+(y-k{)}^2=a^2$, the equation of the tangent at point $(x_1,y_1)$ is given by
$(x_1-h)(x-x_1)+(y_1-k)(y-y_1)=0$

Answer 1

Given the equation of the circle is $(x-h{)}^2+(y-k{)}^2=a^2$.....(1).

Now differentiating both sides of (1) with respect to $x$ we get,

$\frac{dy}{dx}=-\frac{(x-h)}{(y-k)}$.

Now equation of tangent to (1) at $(x_1,y_1)$ is

$(y-y_1)={\frac{dy}{dx}|}_{(x_1,y_1)}(x-x_1)$

or, $(y-y_1)=-\frac{(x_1-h)}{(y_1-k)}(x-x_1)$

or, $(x-x_1)(x_1-h)+(y-y_1)(y_1-k)=0$