The equation of the tangent to the circle x^2 | vedclass

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

Question

(B) The given line is $y = mx + c$. The slope of this line is $m$. Any line perpendicular to this line will have a slope of $-1/m$. Let the equation o

Vedclass · Accepted Answer

$x + my = \pm a\sqrt{1 + m^2}$

Vedclass · Answer

(B) The given line is $y = mx + c$. The slope of this line is $m$. Any line perpendicular to this line will have a slope of $-1/m$. Let the equation of the tangent be $y = -\frac{1}{m}x + k$,which can be rewritten as $x + my - mk = 0$. The condition for a line $Ax + By + C = 0$ to be a tangent to the circle $x^2 + y^2 = a^2$ is that the perpendicular distance from the center $(0, 0)$ to the line equals the radius $a$. Thus,$\frac{|-mk|}{\sqrt{1^2 + m^2}} = a$. $|mk| = a\sqrt{1 + m^2}$,so $mk = \pm a\sqrt{1 + m^2}$. Substituting $mk$ back into the equation $x + my = mk$,we get $x + my = \pm a\sqrt{1 + m^2}$.

The equation of the tangent to the circle $x^2 + y^2 = a^2$ which is perpendicular to the straight line $y = mx + c$ is:

Similar Questions

Let the tangent to the circle $x^{2}+y^{2}=25$ at the point $R(3,4)$ meet the $x$-axis and $y$-axis at points $P$ and $Q$,respectively. If $r$ is the radius of the circle passing through the origin $O$ and having its centre at the incentre of the triangle $OPQ$,then $r^{2}$ is equal to

If the tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+C=0$,then $C=$

The equation of the normal at $t=\frac{\pi}{2}$ to the curve $x=2 \sin t, y=2 \cos t$ is

If the line $4x - 3y + p = 0$ $(p + 3 > 0)$ touches the circle $x^2 + y^2 - 4x + 6y + 4 = 0$ at the point $(h, k)$,then $h - 2k = . . . . . .$

$x-2y-6=0$ is a normal to the circle $x^2+y^2+2gx+2fy-8=0$. If the line $y=2$ touches this circle,then the radius of the circle can be

Vedclass Test Series

Exam Paper Generator

Online Exam Module