The equation of the diameter of the circle x^ | vedclass

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

Question

(C) The given circle is $x^2 + y^2 + 2x - 4y - 11 = 0$. The center of this circle is $(-g, -f) = (-1, 2)$.The diameter that bisects a chord is perpend

Vedclass · Accepted Answer

$x + 2y - 3 = 0$

Vedclass · Answer

(C) The given circle is $x^2 + y^2 + 2x - 4y - 11 = 0$. The center of this circle is $(-g, -f) = (-1, 2)$.The diameter that bisects a chord is perpendicular to that chord.The given line is $2x - y + 3 = 0$,which has a slope of $m_1 = 2$.The slope of the diameter perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{2}$.The equation of the line passing through the center $(-1, 2)$ with slope $-\frac{1}{2}$ is given by $y - 2 = -\frac{1}{2}(x + 1)$.Multiplying by $2$,we get $2y - 4 = -x - 1$.Rearranging the terms,we get $x + 2y - 3 = 0$.

The equation of the diameter of the circle $x^2 + y^2 + 2x - 4y - 11 = 0$ which bisects the chords intercepted on the line $2x - y + 3 = 0$ is

Similar Questions

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. If $P(-5, 6)$ is a point such that $PA=PB=10$,then the equation of line $l$ is:

If $(1, a)$ and $(b, 2)$ are conjugate points with respect to the circle $x^2+y^2=25$,then $4a+2b=$

Find the equation of the diameter of the circle $x^2 + y^2 - 4x + 2y - 11 = 0$ which corresponds to the system of parallel chords $x - 2y + c = 0$.

Let the centre of the circle $S=0$ lie on the line $x+y-5=0$ and also lie in the first quadrant. If this circle touches both the lines $x-2=0$ and $y-5=0$,then the area of the circle is

The lengths of the tangents from the point $(1,2)$ to the circles $x^2+y^2+x+y-4=0$ and $3x^2+3y^2-x-y-k=0$ are in the ratio $4:3$. Then the value of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module