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

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(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

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

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

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

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:

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