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 $x^2+y^2-4x+6y-12=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=3, c=-12$. The center of the circle

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) The equation of the circle is $x^2+y^2-4x+6y-12=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=3, c=-12$. The center of the circle is $(-g, -f) = (2, -3)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-2)^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$. The perpendicular distance $d$ from the center $(2, -3)$ to the line $4x+3y+1=0$ is given by $d = \frac{|4(2)+3(-3)+1|}{\sqrt{4^2+3^2}} = \frac{|8-9+1|}{\sqrt{16+9}} = \frac{0}{5} = 0$. Since the perpendicular distance $d=0$,the line passes through the center of the circle,meaning the chord is a diameter. The length of the chord is $2 	imes \sqrt{r^2-d^2} = 2 	imes \sqrt{5^2-0^2} = 2 	imes 5 = 10$.

The length of the chord intercepted by the circle $x^2+y^2-4x+6y-12=0$ on the line $4x+3y+1=0$ is

Similar Questions

The square of the length of the tangent from the point $(3, -4)$ to the circle $x^2 + y^2 - 4x - 6y + 3 = 0$ is ....

From a point $P$ on the line $4x - 3y = 6$,two tangents are drawn to the circle $x^2 + y^2 - 6x - 4y + 4 = 0$. If the angle between these tangents is $\tan^{-1}\left(\frac{24}{7}\right)$,then $P$ can be:

The radius of a circle whose center is $(2, 1)$ and one of its chords is a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$ is:

The angle subtended by the chord $x+y-1=0$ of the circle $x^2+y^2=1$ at the origin is

Which of the following statements are true and which are false? In each case,give a valid reason for your answer.
$q$: The centre of a circle bisects each chord of the circle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module