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

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Question

(B) The equation of the circle is $x^2 + y^2 = a^2$,which has center $(0, 0)$ and radius $a$.Any line parallel to $y = mx + c$ is of the form $y = mx

Vedclass · Accepted Answer

$y = mx \pm a\sqrt{1 + m^2}$

Vedclass · Answer

(B) The equation of the circle is $x^2 + y^2 = a^2$,which has center $(0, 0)$ and radius $a$.Any line parallel to $y = mx + c$ is of the form $y = mx + k$.For this line to be a tangent to the circle,the perpendicular distance from the center $(0, 0)$ to the line $mx - y + k = 0$ must be equal to the radius $a$.The distance formula is $d = \frac{|m(0) - (0) + k|}{\sqrt{m^2 + (-1)^2}} = a$.This simplifies to $\frac{|k|}{\sqrt{m^2 + 1}} = a$,which gives $|k| = a\sqrt{1 + m^2}$.Thus,$k = \pm a\sqrt{1 + m^2}$.Substituting $k$ back into the line equation,we get $y = mx \pm a\sqrt{1 + m^2}$.

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

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