TS EAMCET 19-July-2022 Shift 2 Solved Paper

Statement I : If the centre of circle lies on $Y$-axis and radius $k$, then equation of circle is $x^2+(y-h)^2=k^2$, where $(0, h)$ is the centre of circle. $2x+2(y-h)\frac{dy}{dx}=0$$\Rightarrow (y-h)\frac{dy}{dx}=-x$$\Rightarrow y-h=-\frac{x}{\frac{dy}{dx}}$ Now, substitute value of $y-h$ in equation of circle, $x^2+\frac{x^2}{\left(\frac{dy}{dx}\right)^2}=k^2$ $\Rightarrow x^2\left(\frac{dy}{dx}\right)^2+x^2=k^2\left(\frac{dy}{dx}\right)^2$ $\Rightarrow (x^2-k^2)\left(\frac{dy}{dx}\right)^2+x^2=0$ $\therefore$ Statement $I$ is true. Statement II : If the centre of circle lies on $X$-axis and circle passes through origin is $(x-h)^2+y^2=h^2$, where centre of circle is $(h, 0)$ and radius is $h$ unit.On differentiate both sides w.r.t. $x$, we get $2(x-h)+2y\frac{dy}{dx}=0$$\Rightarrow x-h+y\frac{dy}{dx}=0$$\Rightarrow h=x+y\frac{dy}{dx}$ Now, substitute the value of $h$ in the equation of circle $\left(x-x-\frac{y\,dy}{dx}\right)^2+y^2=\left(x+\frac{y\,dy}{dx}\right)^2$ $=x^2-y^2+(2xy)\frac{dy}{dx}=0$$\therefore$ Statement II is also true.