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

(B) The equation of the circle is $x^2+y^2-4x-10y+25=0$.Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get the center $C = (-g, -f)

Vedclass · Accepted Answer

$x+3y=7$

Vedclass · Answer

(B) The equation of the circle is $x^2+y^2-4x-10y+25=0$.Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get the center $C = (-g, -f) = (2, 5)$.Let $M(1, 2)$ be the midpoint of the chord $AB$.The chord is perpendicular to the radius $CM$ at the point $M$.The slope of $CM$ is $m_{CM} = \frac{2-5}{1-2} = \frac{-3}{-1} = 3$.Since the chord $AB$ is perpendicular to $CM$,the slope of the chord $AB$ is $m_{AB} = -\frac{1}{m_{CM}} = -\frac{1}{3}$.The equation of the chord passing through $M(1, 2)$ with slope $-\frac{1}{3}$ is given by:$y - 2 = -\frac{1}{3}(x - 1)$$3(y - 2) = -(x - 1)$$3y - 6 = -x + 1$$x + 3y = 7$.

The equation of the chord of the circle $x^2+y^2-4x-10y+25=0$ having its midpoint at $(1,2)$ is

Similar Questions

The equation of a diameter of the circle $x^2 + y^2 - 6x + 2y = 0$ passing through the origin is:

Suppose $C_1$ and $C_2$ are two circles having no common points,then

Find the equation of the pair of straight lines parallel to the $x$-axis and touching the circle $x^2 + y^2 - 6x - 4y - 12 = 0$.

Suppose we have two circles of radius $2$ each in the plane such that the distance between their centers is $2 \sqrt{3}$. The area of the region common to both circles lies between

Let $XY$ be the diameter of a semi-circle with center $O$. Let $A$ be a variable point on the semi-circle and $B$ another point on the semi-circle such that $AB$ is parallel to $XY$. The value of $\angle BOY$ for which the inradius of $\triangle AOB$ is maximum,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module