TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Given, system of family of circles is $x^{2}+y^{2}+2 f y+\lambda(x^{2}+y^{2}+2 g x+k) \;\; =0$ $\Rightarrow \;\; (1+\lambda) x^{2}+(1+\lambda) y^{2}+2 g \lambda x+2 f y+\lambda k \;\; =0$ $\Rightarrow \;\; x^{2}+y^{2}+2 \cdot \left(\frac{g \lambda}{1+\lambda}\right) x+2 \left(\frac{f}{1+\lambda}\right) y+\frac{\lambda k}{1+\lambda}=0$ For point circle, radius should be zero. $\; \therefore \;\; \frac{g^{2} \lambda^{2}}{(1+\lambda)^{2}}+\frac{f^{2}}{(1+\lambda)^{2}}=\frac{\lambda k}{1+\lambda}$ $\Rightarrow\;g^{2} \lambda^{2}+f^{2}=k(\lambda+\lambda^{2})$ $\Rightarrow (g^{2}-k) \lambda^{2}-k \lambda+f^{2}=0$ . . . (i) Let the quadratic Eq. (i) has roots $\lambda_{1}$ and $\lambda_{2}$, then the coordinates of point circles are $c_{1}\left(\frac{-g \lambda_{1}}{1+\lambda_{1}}, \frac{f}{1+\lambda_{1}}\right)$ and $c_{2}\left(\frac{-g \lambda_{2}}{1+\lambda_{2}}, \frac{f}{1+\lambda_{2}}\right)$ $\because$ Lines $O C_{1}$ and $O C_{2}$ are perpendicular to each other. So, $\frac{\frac{f}{1+\lambda_{2}}}{\frac{g \lambda_{2}}{1+\lambda_{2}}} \times \frac{\frac{f}{1+\lambda_{1}}}{\frac{g \lambda_{1}}{1+\lambda_{1}}}=-1$ . . . (ii) $\because$ Product of roots of Eq. (i), we get $\lambda_{1} \lambda_{2}=\frac{f^{2}}{g^{2}-k}$ . . . (iii) From Eqs. (ii) and (iii), we get $\Rightarrow \;\; g^{2}=-g^{2}+k$ $\Rightarrow 2 g^{2}=k \Rightarrow g^{2}=\frac{k}{2}$