Q91 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics
The equations of the tangents drawn from the origin to the circle $x^2+y^2+2 g x+2 f y+g^2=0$ are
- A$x=0,\left(g^2+f^2\right) x-2 g f y=0$
- B$x=0,\left(g^2-f^2\right) x-2 g f y=0$
- Correct answerC$y=0,\left(g^2-f^2\right) y-2 g f x=0$
- D$y=0,\left(g^2+f^2\right) y-2 g f x=0$
Solution
Equation of tangent drawn from origin, ' $O$ ' of the circle is $x^2+y^2+2 g x+2 f y+g^2=0$ or $T^2=S S_1$
$\begin{aligned}
\therefore & \left(0+0+2 g\left(\frac{x+0}{2}\right)+2 f \frac{(y+0)}{2}+g^2\right)^2 \\
& =\left(x^2+y^2+2 g x+2 f y+g^2\right)\left(g^2\right) \\
\Rightarrow & \left(g x+f y+g^2\right)^2=g^2\left(x^2+y^2\right)+2 g^3 x+2 f g^2 y+g^4 \\
\Rightarrow & g^2 x^2+f^2 y^2+g^4+2 g^3 x+2 g^2 f y+2 g f x y \\
& =g^2 x^2+g^2 y^2+2 g^3 x+2 f g^2 y+g^4 \\
\Rightarrow \quad & f^2 y^2+2 g^2 f y+2 g f x y=g^2 y^2+2 f g^2 y \\
\Rightarrow & y^2\left(g^2-f^2\right)-2 g f y x=0 \Rightarrow y\left[\left(g^2-f^2\right) y-2 g f x\right]=0 \\
\Rightarrow & y=0,\left(g^2-f^2\right) y-2 g f x=0
\end{aligned}$
Hence, the correct answer is option: 'C'.