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

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(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:

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