TS EAMCET 19-July-2022 Shift 2 Solved Paper

Given, $O(0,0)$ and $A(1,0)$ are the centres of 2 units circles $C_1$ and $C_3$.The equation of circle passes through $O(0,0)$ and $A(1,0)$ is The equation of circle passes through $O(0,0)$ and $A(1,0)$ is If it represents another unit circle $C_3$, its radius $=1$$\;\text{ Here, }\; g=-\frac{1}{2},\; f=\frac{\lambda}{2},\; c=0$ $\Rightarrow r=\sqrt{\frac{1}{4}+\frac{\lambda^2}{4}-0} \Rightarrow 1=\frac{1}{4}+\frac{\lambda^2}{4}$$\Rightarrow \lambda^2=4-1 \Rightarrow \lambda=\pm \sqrt{3}$ As the centre of $C_3$ lies above $X$-axis, $f$ is negative, so $\lambda$ is also negative.$\therefore \;\; \lambda=-\sqrt{3}$ Equation of circle is$x^2+y^2-x-\sqrt{3} y=0$ Since, $C_1$ and $C_3$ intersect and are of unit radius, their common tangent are parallel to the line joining the centres $(0,0)$ and $\left(\frac{1}{2},\; \frac{\sqrt{3}}{2}\right)$. Let the equation of a common tangent be$\sqrt{3} x-y+k=0$ It will touch $C_1$ if $\frac{|k|}{\sqrt{(\sqrt{3})^2+(-1)^2}}=1$$\;k=\pm 2$$\;$ $\therefore$ The required positive tangent make positive intercept on $Y$-axis and negative on $X$-axis and hence, the required equation is$\sqrt{3} x-y+2=0$