For what possible value of p is the line x co | vedclass

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

Question

(C) The given equation of the circle is $x^2 + y^2 - 2qx \cos \alpha - 2qy \sin \alpha = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2

Vedclass · Accepted Answer

$0$ and $2q$

Vedclass · Answer

(C) The given equation of the circle is $x^2 + y^2 - 2qx \cos \alpha - 2qy \sin \alpha = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -q \cos \alpha$ and $f = -q \sin \alpha$. The center of the circle is $(-g, -f) = (q \cos \alpha, q \sin \alpha)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{q^2 \cos^2 \alpha + q^2 \sin^2 \alpha} = \sqrt{q^2(1)} = |q|$. For the line $x \cos \alpha + y \sin \alpha - p = 0$ to be a tangent to the circle,the perpendicular distance from the center to the line must be equal to the radius $r$. Distance $d = \frac{|(q \cos \alpha)(\cos \alpha) + (q \sin \alpha)(\sin \alpha) - p|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = |q(\cos^2 \alpha + \sin^2 \alpha) - p| = |q - p|$. Setting $d = r$,we get $|q - p| = |q|$. This implies $q - p = q$ or $q - p = -q$. Therefore,$p = 0$ or $p = 2q$.

For what possible value of $p$ is the line $x \cos \alpha + y \sin \alpha = p$ a tangent to the circle $x^2 + y^2 - 2qx \cos \alpha - 2qy \sin \alpha = 0$?

Similar Questions

The equation of the tangent to the circle $x^2 + y^2 - 22x - 4y + 25 = 0$ which is perpendicular to the line $5x + 12y + 8 = 0$ is:

The point on the curve ${y^2} = 2(x - 3)$ at which the normal is parallel to the line $y - 2x + 1 = 0$ is

Find the equation of the circle having normals $(x-1)(y-2)=0$ and a tangent $3x+4y=6$.

If the normals of the parabola $y^2 = 4x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^2 + (y + 2)^2 = r^2$,then the value of $r^2$ is

The normal at a point $\theta$ to the curve $x=a(1+\cos \theta), y=a \sin \theta$ always passes through the fixed point

Vedclass Test Series

Exam Paper Generator

Online Exam Module