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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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)

Vedclass · Accepted Answer

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

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

Similar Questions

Let $P$ be any point on the circle $x^2+y^2-2x-1=0$ and $C$ be its centre. Let $AB$ be the chord of contact of $P$ with respect to the circle $x^2+y^2-2x=0$. Then the locus of the circumcentre of the triangle $CAB$ is

Tangents are drawn from the point $(-1, -4)$ to the circle $x^2 + y^2 - 2x + 4y + 1 = 0$. The length of the corresponding chord of contact is:

The length of the chord of contact of tangents drawn from the point $(4,4)$ to the circle $x^2 + y^2 - 2x - 2y - 7 = 0$ is: (in $\sqrt{2}$)

If the common chord of the circles $x^2+y^2-2x+2y+1=0$ and $x^2+y^2-2x-2y-2=0$ is the diameter of a circle $S$,then the centre of the circle $S$ is

$A$ pair of tangents is drawn from every point on the line $3x + 4y = 12$ to the circle $x^2 + y^2 = 4$. Their variable chord of contact always passes through a fixed point whose coordinates are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module