Show that, if $x^2+y^2=1$, then the point $(\sqrt{x^2+y^2},\sqrt{1-x^2-y^2})$ is at a distance $1$ unit from the origin.

Q. The equation of the circle in the first quadrant touching each coordinate axes at a distance of one unit from the origin is
(a) x² + y² – 2x – 2y + 1 = 0
(b) x² + y² – 2x – 2y – 1 = 0
(c) x² + y² – 2x – 2y = 0
(d) x² + y² – 2x + 2y – 1 = 0

Q.

Tangents are drawn from the origin to the curve y = sin x, then their point of contact lie on the curve

Q. If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$ then show that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$.

Q. If ${\cos }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)={\tan }^{-1}\left(a\right)$, then show that $\frac{dy}{dx}=\frac{y}{x}.$