The length of the chord of contact of the point $(2,1)$ with respect to the circle $x^2+y^2+4 x+2 y+1=0$ is
The length of the chord of contact of the point $(2,1)$ with respect to the circle $x^2+y^2+4 x+2 y+1=0$ is
- $\frac{8}{\sqrt{5}}$
- $\frac{4}{\sqrt{5}}$
- $\frac{4 \sqrt{6}}{\sqrt{5}}$
- $\frac{2 \sqrt{6}}{\sqrt{5}}$
Solution
The given equation of circle is:
$\begin{aligned} & x^2+y^2+4 x+2 y+1=0 \\ & C \equiv(-2,-1), r=\sqrt{4+1-1}=2\end{aligned}$
The equation of chord of contact of point $(2,1)$ is:
$\begin{aligned} & 2 x+1 y+\frac{4(x+2)}{2}+\frac{2(y+1)}{2}+1=0 \\ & \Rightarrow \quad 2 x+y+2(x+2)+(y+1)+1=0 \\ & \Rightarrow \quad 2 x+y+3=0\end{aligned}$
$\begin{aligned} & \mathrm{CM}=\left|\frac{-4-1+3}{\sqrt{5}}\right|=\frac{2}{\sqrt{5}} \\ & \mathrm{PM}=\sqrt{\mathrm{CP}^2-\mathrm{CM}^2} \\ & =\sqrt{4-\frac{4}{5}}\end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{PM}=\frac{4}{\sqrt{5}} \\ & \Rightarrow \mathrm{PQ}=2 \mathrm{PM}=2 \times \frac{4}{\sqrt{5}}=\frac{8}{\sqrt{5}}\end{aligned}$
Asked in: AP EAMCET 2023 (18 May Shift 1)
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