The equation of tangents to the circle $x^2+y^2=4$ which are parallel to $x+2 y+3=0$ are

The equation of tangents to the circle $x^2+y^2=4$ which are parallel to $x+2 y+3=0$ are
  1. $x+2 y= \pm 2 \sqrt{5}$
  2. $x+2 y= \pm 2 \sqrt{3}$
  3. $x-2 y= \pm 2$
  4. $x-2 y= \pm 2 \sqrt{5}$

Solution

Equation of tangent to the circle $x^2+y^2=r^2$ having slope $m$ is $\begin{aligned} & \Rightarrow y=-\frac{1}{2} x \pm 2 \sqrt{1+\left(\frac{-1}{2}\right)^2} \quad\left[\text { Here } \mathrm{r}=2 \text { and } \mathrm{m}=\frac{-1}{2}\right] \\ & \Rightarrow \mathrm{y}=\frac{-\mathrm{x}}{2} \pm 2 \times \frac{\sqrt{5}}{2} \\ & \Rightarrow 2 \mathrm{y}=-\mathrm{x} \pm 2 \sqrt{5} \\ & \Rightarrow \mathrm{x}+2 \mathrm{y}= \pm 2 \sqrt{5}\end{aligned}$

Asked in: MHT CET 2022 (05 Aug Shift 2)

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