If the line (x + g) cos theta + (y + f) sin t | vedclass

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

Question

(C) The given circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Its center is $C(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$.The given line is $(x + g) \co

Vedclass · Accepted Answer

$g^2 + f^2 = k^2 + c$

Vedclass · Answer

(C) The given circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Its center is $C(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$.The given line is $(x + g) \cos 	heta + (y + f) \sin 	heta = k$,which can be rewritten as $x \cos 	heta + y \sin 	heta + (g \cos 	heta + f \sin 	heta - k) = 0$.If the line is tangent to the circle,the perpendicular distance from the center $C(-g, -f)$ to the line must equal the radius $r$.$r = \left| \frac{(-g) \cos 	heta + (-f) \sin 	heta + g \cos 	heta + f \sin 	heta - k}{\sqrt{\cos^2 	heta + \sin^2 	heta}} ight|$.Simplifying the numerator,we get $r = \left| \frac{-k}{1} ight| = |k|$.Thus,$\sqrt{g^2 + f^2 - c} = |k|$.Squaring both sides,we get $g^2 + f^2 - c = k^2$,which simplifies to $g^2 + f^2 = k^2 + c$.

If the line $(x + g) \cos \theta + (y + f) \sin \theta = k$ is tangent to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$,then:

Similar Questions

If $y=3x$ is a tangent to a circle with centre $(1,1)$,then the other tangent drawn through $(0,0)$ to the circle is

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

The equation of the circle passing through the point $(-1, -3)$ and touching the line $4x + 3y - 12 = 0$ at the point $(3, 0)$ is

If $2x - 3y + 5 = 0$ and $4x - 5y + 7 = 0$ are the equations of the normals drawn to a circle and $(2, 5)$ is a point on the given circle,then the radius of the circle is

The tangent to the circle $x^2 + y^2 = 5$ at the point $(1, -2)$ also touches the circle $x^2 + y^2 - 8x + 6y + 20 = 0$. Find its point of contact.

Vedclass Test Series

Exam Paper Generator

Online Exam Module