[tex] \small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}[/tex][tex] \rule{300pt}{0.1pt}[/tex][tex]\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex][tex] \rule{300pt}{0.1pt}[/tex]
[tex] \therefore \quad[/tex] We known that the equation of tangent to the circle [tex]\tt x^{2}+y^{2}+2 g x+2 f y+c=0[/tex] at point [tex]\tt \left(x_{1}, y_{1}\right)[/tex] on it, is given by
[tex] \tt\[ x x_{1}+y y_{1}+g\left(x+x_{1}\right)+f\left(y+y_{1}\right)+c=0 \] \\ [/tex]
In this case, we have [tex]\tt \left(x, y_{1}\right)=(0,0)[/tex]
[tex] \text{\( \tt 2 g=-3: g=\dfrac{-3}{2} \) and \( \tt 2 f=2 \quad \therefore f=1 \) and c=0}[/tex]
[tex] \therefore \quad[/tex] The equation of tangent is
[tex] \small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}[/tex][tex] \rule{300pt}{0.1pt}[/tex][tex]\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex][tex] \rule{300pt}{0.1pt}[/tex]
[tex] \therefore \quad[/tex] We known that the equation of tangent to the circle [tex]\tt x^{2}+y^{2}+2 g x+2 f y+c=0[/tex] at point [tex]\tt \left(x_{1}, y_{1}\right)[/tex] on it, is given by
[tex] \tt\[ x x_{1}+y y_{1}+g\left(x+x_{1}\right)+f\left(y+y_{1}\right)+c=0 \] \\ [/tex]
In this case, we have [tex]\tt \left(x, y_{1}\right)=(0,0)[/tex]
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[tex] \small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}[/tex][tex] \rule{300pt}{0.1pt}[/tex][tex]\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex][tex] \rule{300pt}{0.1pt}[/tex]
[tex] \therefore \quad[/tex] We known that the equation of tangent to the circle [tex]\tt x^{2}+y^{2}+2 g x+2 f y+c=0[/tex] at point [tex]\tt \left(x_{1}, y_{1}\right)[/tex] on it, is given by
[tex] \tt\[ x x_{1}+y y_{1}+g\left(x+x_{1}\right)+f\left(y+y_{1}\right)+c=0 \] \\ [/tex]
In this case, we have [tex]\tt \left(x, y_{1}\right)=(0,0)[/tex]
[tex] \text{\( \tt 2 g=-3: g=\dfrac{-3}{2} \) and \( \tt 2 f=2 \quad \therefore f=1 \) and c=0}[/tex]
[tex] \therefore \quad[/tex] The equation of tangent is
[tex]\[ \begin{aligned} & \tt x(0)+y(0)+\left(\frac{-3}{2}\right)(x+0)+(1)(y+0)+0=0 \\ \\ & \tt\frac{-3}{2} x+y=0 \\ \\ \therefore \quad & \tt 3 x-2 y=0 \end{aligned} \][/tex]
[tex] \small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}[/tex][tex] \rule{300pt}{0.1pt}[/tex][tex]\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex][tex] \rule{300pt}{0.1pt}[/tex]
[tex] \therefore \quad[/tex] We known that the equation of tangent to the circle [tex]\tt x^{2}+y^{2}+2 g x+2 f y+c=0[/tex] at point [tex]\tt \left(x_{1}, y_{1}\right)[/tex] on it, is given by
[tex] \tt\[ x x_{1}+y y_{1}+g\left(x+x_{1}\right)+f\left(y+y_{1}\right)+c=0 \] \\ [/tex]
In this case, we have [tex]\tt \left(x, y_{1}\right)=(0,0)[/tex]
[tex] \text{\( \tt 2 g=-3: g=\dfrac{-3}{2} \) and \( \tt 2 f=2 \quad \therefore f=1 \)}[/tex]
and c=0 [tex] \therefore \quad[/tex] The equation of tangent is
[tex]\[ \begin{aligned} & \tt x(0)+y(0)+\left(\frac{-3}{2}\right)(x+0)+(1)(y+0)+0=0 \\ \\ & \tt\frac{-3}{2} x+y=0 \\ \\ \therefore \quad & \tt 3 x-2 y=0 \end{aligned} \][/tex]