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

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

Question

(C) The given circle is $x^2 + y^2 - 2x = 0$. Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$ and $f = 0$. The c

Vedclass · Accepted Answer

$x + 2y - 1 = 0$

Vedclass · Answer

(C) The given circle is $x^2 + y^2 - 2x = 0$. Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$ and $f = 0$. The centre of the circle is $(-g, -f) = (1, 0)$. Any line parallel to $x + 2y = 3$ is of the form $x + 2y + \lambda = 0$. Since the normal to a circle always passes through its centre,the line $x + 2y + \lambda = 0$ must pass through $(1, 0)$. Substituting the point $(1, 0)$ into the equation: $1 + 2(0) + \lambda = 0$,which gives $\lambda = -1$. Thus,the equation of the normal is $x + 2y - 1 = 0$.

The equation of the normal to the circle $x^2 + y^2 - 2x = 0$ parallel to the line $x + 2y = 3$ is

Similar Questions

If $y = c$ is a tangent to the circle $x^2 + y^2 - 2x + 2y - 2 = 0$ at the point $(1, 1)$,then the value of $c$ is:

The area of the triangle formed by the tangent and the normal drawn at $(1, \sqrt{3})$ to the circle $x^2 + y^2 = 4$ and the positive $x$-axis is:

If the equation of one tangent to the circle with center at $(2, -1)$ from the origin is $3x + y = 0$, then the equation of the other tangent through the origin is

The area of the region (in sq. unit) bounded by the $x$-axis, the tangent, and the normal to the circle $x^2 + y^2 = 4$ drawn at the point $(1, \sqrt{3})$ is

The equation of the normal to the circle $x^2+y^2+6x+4y-3=0$ at the point $(1, -2)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module