The length of the chord of contact of the poi | vedclass

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(A) The given equation of the circle is $x^2+y^2+4x+2y+1=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=1, c=1$. The center $C$ is $(-g, -f)

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$\frac{8}{\sqrt{5}}$

Vedclass · Answer

(A) The given equation of the circle is $x^2+y^2+4x+2y+1=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=1, c=1$. The center $C$ is $(-g, -f) = (-2, -1)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+1-1} = 2$. The equation of the chord of contact of the point $(x_1, y_1) = (2, 1)$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$. Substituting the values: $2x + 1y + 2(x+2) + 1(y+1) + 1 = 0$. $2x + y + 2x + 4 + y + 1 + 1 = 0$ $\Rightarrow 4x + 2y + 6 = 0$ $\Rightarrow 2x + y + 3 = 0$. Let $CM$ be the perpendicular distance from the center $(-2, -1)$ to the chord $2x + y + 3 = 0$. $CM = \frac{|2(-2) + 1(-1) + 3|}{\sqrt{2^2 + 1^2}} = \frac{|-4 - 1 + 3|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$. Let $PM$ be half the length of the chord. In $	riangle CPM$,$PM = \sqrt{r^2 - CM^2} = \sqrt{2^2 - (\frac{2}{\sqrt{5}})^2} = \sqrt{4 - \frac{4}{5}} = \sqrt{\frac{16}{5}} = \frac{4}{\sqrt{5}}$. The length of the chord of contact $PQ = 2 	imes PM = 2 	imes \frac{4}{\sqrt{5}} = \frac{8}{\sqrt{5}}$.

The length of the chord of contact of the point $(2,1)$ with respect to the circle $x^2+y^2+4x+2y+1=0$ is

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