The equation of the chord of the circle x^2+y | vedclass

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

Question

(D) The equation of a circle is given by $x^2+y^2-4x=0$.Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$, we get $g=-2$, $f=0$, and $c=0$.Th

Vedclass · Accepted Answer

$x=1$

Vedclass · Answer

(D) The equation of a circle is given by $x^2+y^2-4x=0$.Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$, we get $g=-2$, $f=0$, and $c=0$.The center of the circle is $(-g, -f) = (2, 0)$.The equation of a chord of a circle with a given midpoint $(x_1, y_1)$ is given by $T = S_1$, where $T = xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c$ and $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.Here, $(x_1, y_1) = (1, 0)$.$T = x(1) + y(0) - 2(x+1) + 0(y+0) + 0 = x - 2x - 2 = -x - 2$.$S_1 = (1)^2 + (0)^2 - 4(1) = 1 - 4 = -3$.Equating $T = S_1$, we get $-x - 2 = -3$, which simplifies to $x = 1$.

The equation of the chord of the circle $x^2+y^2-4x=0$ whose midpoint is $(1,0)$ is

Similar Questions

Which of the following lines is a diameter of the circle $x^2 + y^2 - 6x - 8y - 9 = 0$?

Consider the circles ${x^2} + {(y - 1)^2} = 9$ and ${(x - 1)^2} + {y^2} = 25$. They are such that:

If $A\left(\frac{\pi}{3}\right)$ and $B\left(\frac{\pi}{6}\right)$ are points on a circle represented in parametric form with center $(0,0)$ and radius $12$,then the length of the chord $AB$ is:

If the circle $x^2+y^2-4x-8y-5=0$ intersects the line $3x-4y-m=0$ in two distinct points,then the number of integral values of '$m$' is

Let $PQ$ be a diameter of the circle $x^{2}+y^{2}=9$. If $\alpha$ and $\beta$ are the lengths of the perpendiculars from $P$ and $Q$ on the straight line $x+y=2$ respectively,then the maximum value of $\alpha \beta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module