The combined equation of the direct common tangents of the circles $x^2+y^2+2 x=0$ and $x^2+y^2-2 y-3=0$ is

• Correct answer
  A

  $x y+x+2 y+2=0$
• B

  $x^2-x y-2 y^2+3 x-6 y=0$
• C

  $2 x^2+5 x y+2 y^2+13 x+14 y+20=0$
• D

  $2 x^2-9 x y+9 y^2+3 x-6 y+1=0$

Solution

For the circle $$ \begin{aligned} & x^2+y^2+2 x=0 \\ & \text { centre }=(-1,0) \\ & \text { Radius }=1 \end{aligned} $$ For the circle $$ \begin{aligned} & x^2+y^2-2 y-3=0 \\ & \text { centre }=(0,1) \\ & \text { Radius }=2 \end{aligned} $$ $$ \mathrm{c}_1 \mathrm{c}_2=\sqrt{(0+1)^2+(1-0)^2}=\sqrt{2} < \mathrm{r}_1+\mathrm{r}_2 $$ In $\triangle \mathrm{ABC}: \sin \theta=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}$ In $\Delta \mathrm{AC}_2 \mathrm{P}: \sin 45^{\circ}=\frac{2}{\mathrm{c}_2 \mathrm{p}}$ $$ \begin{aligned} & \Rightarrow \frac{1}{\sqrt{2}}=\frac{2}{\mathrm{c}_2 \mathrm{p}} \Rightarrow \mathrm{c}_2 \mathrm{p}=2 \sqrt{2} \\ & \text { and } \mathrm{c}_1 \mathrm{p}=2 \sqrt{2}-\sqrt{2}=\sqrt{2} \\ & \therefore \mathrm{c}_1 \mathrm{p}=\mathrm{c}_1 \mathrm{c}_2 \text { so, } \mathrm{c}_1 \text { is mid point of } \mathrm{c}_2 \mathrm{p} \\ & \therefore \mathrm{p}(-2,-1) \end{aligned} $$ Slope of $\mathrm{c}_1 \mathrm{c}_2=\frac{1-0}{0+1}=1$ $\tan 45^{\circ}=\left|\frac{1-\mathrm{m}}{1+\mathrm{m}}\right| \Rightarrow 1=\frac{1-\mathrm{m}}{1+\mathrm{m}}$ $$ \mathrm{m}=0 \text { i.e. } \mathrm{m}_1=0 $$ $$ \therefore \angle \mathrm{p}=90^{\circ} \text { so }_2=\frac{1}{0} $$ Equations of common tangents are: $$ \mathrm{x}+2=0 \text { and } \mathrm{y}+1=0 $$ Combined equations is $$ (x+2)(y+1)=0 \Rightarrow x y+x+2 y+2=0 $$

Hence, the correct answer is option: 'A'.