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

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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:

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