If the line y cos alpha = x sin alpha + a cos | vedclass

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(B) The given line is $y \cos \alpha = x \sin \alpha + a \cos \alpha$.Dividing by $\cos \alpha$ (assuming $\cos \alpha 
eq 0$),we get $y = x 	an \al

Vedclass · Accepted Answer

$\cos^2 \alpha = 1$

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(B) The given line is $y \cos \alpha = x \sin \alpha + a \cos \alpha$.Dividing by $\cos \alpha$ (assuming $\cos \alpha 
eq 0$),we get $y = x 	an \alpha + a$.This is in the form $y = mx + c$,where $m = 	an \alpha$ and $c = a$.The condition for the line $y = mx + c$ to be a tangent to the circle $x^2 + y^2 = a^2$ is $c^2 = a^2(1 + m^2)$.Substituting the values,we get $a^2 = a^2(1 + 	an^2 \alpha)$.Since $a^2 
eq 0$,we have $1 = 1 + 	an^2 \alpha$,which implies $	an^2 \alpha = 0$.Alternatively,using the perpendicular distance from the center $(0, 0)$ to the line $x \sin \alpha - y \cos \alpha + a \cos \alpha = 0$ being equal to the radius $a$:$\frac{|0 - 0 + a \cos \alpha|}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} = a$$|a \cos \alpha| = a$$|\cos \alpha| = 1$$\cos^2 \alpha = 1$.

If the line $y \cos \alpha = x \sin \alpha + a \cos \alpha$ is a tangent to the circle $x^2 + y^2 = a^2$,then

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