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

(A) The equation of the circle is $x^2 + y^2 = 25$,which has center $(0, 0)$ and radius $r = 5$.The slope of the tangent is $m = 	an(60^{\circ}) = \s

Vedclass · Accepted Answer

$y = \sqrt{3}x \pm 10$

Vedclass · Answer

(A) The equation of the circle is $x^2 + y^2 = 25$,which has center $(0, 0)$ and radius $r = 5$.The slope of the tangent is $m = 	an(60^{\circ}) = \sqrt{3}$.The equation of a line with slope $m$ tangent to the circle $x^2 + y^2 = r^2$ is given by $y = mx \pm r\sqrt{1 + m^2}$.Substituting $m = \sqrt{3}$ and $r = 5$:$y = \sqrt{3}x \pm 5\sqrt{1 + (\sqrt{3})^2}$$y = \sqrt{3}x \pm 5\sqrt{1 + 3}$$y = \sqrt{3}x \pm 5\sqrt{4}$$y = \sqrt{3}x \pm 5(2)$$y = \sqrt{3}x \pm 10$.

The equation of the tangent to the circle $x^2 + y^2 = 25$ which is inclined at an angle of $60^{\circ}$ with the $x$-axis is:

Similar Questions

At which point does the line $3x + 4y = 25$ touch the circle $x^2 + y^2 = 25$?

If the line $y = mx + C$ is a tangent to the circle $x^2 + y^2 = 16$,then $m =$

The angle between the pair of tangents drawn from $(1,1)$ to the circle $x^2+y^2+4x+4y-1=0$ 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

The equations of the tangents to the circle $x^2 + y^2 = 4$,which are parallel to $x + 2y + 3 = 0$,are

Vedclass Test Series

Exam Paper Generator

Online Exam Module