The length of the chord intercepted by the ci | vedclass

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

Question

(C) The equation of the circle is $x^2+y^2+2x+4y-20=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=1, f=2, c=-20$. The center of the circle is $C(

Vedclass · Accepted Answer

$\frac{8}{5} \sqrt{21}$

Vedclass · Answer

(C) The equation of the circle is $x^2+y^2+2x+4y-20=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=1, f=2, c=-20$. The center of the circle is $C(-g, -f) = (-1, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{1^2+2^2-(-20)} = \sqrt{25} = 5$. The perpendicular distance $d$ from the center $(-1, -2)$ to the line $3x+4y-6=0$ is given by $d = \frac{|3(-1)+4(-2)-6|}{\sqrt{3^2+4^2}} = \frac{|-3-8-6|}{5} = \frac{17}{5}$. The length of the chord is $2\sqrt{r^2-d^2} = 2\sqrt{5^2 - (\frac{17}{5})^2} = 2\sqrt{25 - \frac{289}{25}} = 2\sqrt{\frac{625-289}{25}} = 2\sqrt{\frac{336}{25}} = 2 	imes \frac{\sqrt{16 	imes 21}}{5} = 2 	imes \frac{4\sqrt{21}}{5} = \frac{8}{5}\sqrt{21}$.

The length of the chord intercepted by the circle $x^2+y^2+2x+4y-20=0$ on the line $3x+4y-6=0$ is

Similar Questions

$A$ line is drawn through a fixed point $P(\alpha, \beta)$ to cut the circle $x^{2}+y^{2}=r^{2}$ at $A$ and $B$. Then $PA \cdot PB$ is equal to

Find a possible equation of a chord of the circle $x^2 + y^2 = 100$ which passes through the point $(1, 7)$ and makes an angle of $\frac{2\pi}{3}$ with the origin.

If a circle and a square have the same perimeter,then

If the ratio of the lengths of the tangents drawn from the point $(1, 2)$ to the circles $x^2 + y^2 + x + y - 4 = 0$ and $3x^2 + 3y^2 - x - y + k = 0$ is $4 : 3$,then $k = \dots$

$A$ circle $S$ touches the $Y$-axis at $(0,3)$ and makes an intercept of length $8$ units on the $X$-axis. If the centre $C$ of the circle $S$ lies in the second quadrant,then the distance of $C$ from the point $(-2,-1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module