The condition that the line x cos alpha + y s | vedclass

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

Question

(C) For a line to touch a circle,the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.The eq

Vedclass · Accepted Answer

${p^2} = {a^2}$

Vedclass · Answer

(C) For a line to touch a circle,the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.The equation of the circle is ${x^2} + {y^2} = {a^2}$,so its center is $(0, 0)$ and its radius is $a$.The perpendicular distance from the center $(0, 0)$ to the line $x \cos \alpha + y \sin \alpha - p = 0$ is given by:$d = \frac{|(0)\cos \alpha + (0)\sin \alpha - p|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}}$Since $\cos^2 \alpha + \sin^2 \alpha = 1$,we have:$d = \frac{|-p|}{\sqrt{1}} = |p|$Setting the distance equal to the radius $(d = a)$:$|p| = a$Squaring both sides,we get:${p^2} = {a^2}$

The condition that the line $x \cos \alpha + y \sin \alpha = p$ may touch the circle ${x^2} + {y^2} = {a^2}$ is

Similar Questions

The tangent to the circle $x^2 + y^2 = 10$ at the point $(3, 1)$ touches the circle $x^2 + y^2 - 2\sqrt{10}x - 20y + k = 0$. Find the value of $k$.

The length of the tangent drawn to the circle $x^2+y^2-2x+4y-11=0$ from the point $(1,3)$ is:

For what value of $m$ does the line $3x + 4y = m$ touch the circle $x^2 + y^2 - 2x - 8 = 0$?

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

Consider a circle $(x-\alpha)^2+(y-\beta)^2=50$,where $\alpha, \beta > 0$. If the circle touches the line $y+x=0$ at the point $P$,whose distance from the origin is $4 \sqrt{2}$,then $(\alpha+\beta)^2$ is equal to................

Vedclass Test Series

Exam Paper Generator

Online Exam Module