If a line passing through the origin touches | vedclass

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

Question

(A) The equation of the line passing through the origin is $y = mx$,which can be written as $mx - y = 0$.The circle is given by $(x - 4)^2 + (y + 5)^2

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The equation of the line passing through the origin is $y = mx$,which can be written as $mx - y = 0$.The circle is given by $(x - 4)^2 + (y + 5)^2 = 25$,which has center $(4, -5)$ and radius $r = 5$.For the line to be tangent to the circle,the perpendicular distance from the center $(4, -5)$ to the line $mx - y = 0$ must be equal to the radius $r = 5$.Using the distance formula: $\frac{|m(4) - (-5)|}{\sqrt{m^2 + (-1)^2}} = 5$.$|4m + 5| = 5\sqrt{m^2 + 1}$.Squaring both sides: $(4m + 5)^2 = 25(m^2 + 1)$.$16m^2 + 40m + 25 = 25m^2 + 25$.$9m^2 - 40m = 0$.$m(9m - 40) = 0$.Thus,$m = 0$ or $m = \frac{40}{9}$.

If a line passing through the origin touches the circle $(x - 4)^2 + (y + 5)^2 = 25$,then its slope is:

Similar Questions

The equation of a line parallel to the line $3x + 4y = 0$ and touching the circle $x^{2} + y^{2} = 9$ in the first quadrant is:

If the straight line $y = mx + c$ touches the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the point $(2, 3)$,then $c =$

If the centre of a circle is $(-6, 8)$ and it passes through the origin,then the equation of its tangent at the origin is

The straight line $x - y - 3 = 0$ touches the circle $x^2 + y^2 - 4x + 6y + 11 = 0$ at the point whose coordinates are

If the length of the tangent drawn from the point $(5, 3)$ to the circle $x^2 + y^2 + 2x + ky + 17 = 0$ is $7$,then the value of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module