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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$\cos^2 \alpha = 1$

Vedclass · Answer

(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

Similar Questions

Let $O$ be the origin and $OP$ and $OQ$ be the tangents to the circle $x^2+y^2-6x+4y+8=0$ at the points $P$ and $Q$ on it. If the circumcircle of the triangle $OPQ$ passes through the point $(\alpha, \frac{1}{2})$,then a value of $\alpha$ is

The equation of the tangent to the circle $x^2+y^2=1$,which is perpendicular to the line $y=mx+1$,is:

Point $M$ moves along the circle $(x - 4)^2 + (y - 8)^2 = 20$. It then breaks away from the circle and moves along a tangent to the circle,passing through the point $(-2, 0)$ on the $x$-axis. The coordinates of the point on the circle at which the moving point broke away can be:

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

The length of the tangent from the point $(4, 5)$ to the circle $x^2 + y^2 + 2x - 6y - 6 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module