The line 3x - 4y = 0 is: | vedclass

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

Question

(B) The given circle is $x^2 + y^2 = 25$,which has its center at the origin $(0, 0)$.The perpendicular distance $d$ from the center $(0, 0)$ to the li

Vedclass · Accepted Answer

$A$ normal to the circle $x^2 + y^2 = 25$.

Vedclass · Answer

(B) The given circle is $x^2 + y^2 = 25$,which has its center at the origin $(0, 0)$.The perpendicular distance $d$ from the center $(0, 0)$ to the line $3x - 4y = 0$ is given by:$d = \frac{|3(0) - 4(0)|}{\sqrt{3^2 + (-4)^2}} = \frac{0}{5} = 0$.Since the perpendicular distance from the center to the line is $0$,the line passes through the center of the circle.Any line passing through the center of a circle is a normal to that circle.Therefore,the line $3x - 4y = 0$ is a normal to the circle $x^2 + y^2 = 25$.

The line $3x - 4y = 0$ is:

Similar Questions

If the line $x+3y=0$ is the tangent at $(0,0)$ to the circle of radius $1$,then the centre of one such circle is

The equation of the normal to the circle $x^2+y^2-2x+4y-5=0$ at the point $(2,1)$ is:

If the equation of the common tangent at the point $(1, -1)$ to the two circles,each of radius $13$,is $12x + 5y - 7 = 0$,then the centers of the two circles are

If the area of the triangle formed by the positive $x-$axis,the normal and the tangent to the circle $(x-2)^{2}+(y-3)^{2}=25$ at the point $(5,7)$ is $A$,then $24A$ is equal to ...... .

If two tangents are drawn from the point $P$ on the circle $x^2+y^2=4$ to the circle $x^2+y^2=1$,where the point $P$ is given by $(\sqrt{2}, \sqrt{2})$,then the slopes of the tangents are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module