Answer
$x=-\dfrac{at}{\sqrt{1+t^2}}$; $y=\dfrac{a}{\sqrt{1+t^2}}$ ; $-\infty \lt t \lt \infty$
Work Step by Step
The equation of the circle is $x^2+y^2=a^2$
Here, $x^2+y^2=a^2 \implies y=\sqrt {a^2-x^2}$
and slope: $t=\dfrac{dy}{dx}=-\dfrac{x}{\sqrt {a^2-x^2}}$
or, $x^2=\dfrac{(at)^2}{1+t^2}$
$x=\dfrac{at}{\sqrt{1+t^2}}$ or $x=-\dfrac{at}{\sqrt{1+t^2}}$
Now, $y=\sqrt {a^2-x^2} \implies \sqrt {a^2-\dfrac{(at)^2}{1+t^2}}=\dfrac{a}{\sqrt{1+t^2}}$
Hence, $x=-\dfrac{at}{\sqrt{1+t^2}}$; $y=\dfrac{a}{\sqrt{1+t^2}}$ ; $-\infty \lt t \lt \infty$
