If a circle with center (0, 0) touches the li | vedclass

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

Question

(C) The radius $r$ of the circle is the perpendicular distance from the center $(0, 0)$ to the line $5x + 12y - 1 = 0$.$r = \frac{|5(0) + 12(0) - 1|}{

Vedclass · Accepted Answer

$169 (x^2 + y^2) = 1$

Vedclass · Answer

(C) The radius $r$ of the circle is the perpendicular distance from the center $(0, 0)$ to the line $5x + 12y - 1 = 0$.$r = \frac{|5(0) + 12(0) - 1|}{\sqrt{5^2 + 12^2}} = \frac{|-1|}{\sqrt{25 + 144}} = \frac{1}{\sqrt{169}} = \frac{1}{13}$.The equation of a circle with center $(0, 0)$ and radius $r$ is $x^2 + y^2 = r^2$.Substituting $r = \frac{1}{13}$,we get $x^2 + y^2 = (\frac{1}{13})^2 = \frac{1}{169}$.Multiplying both sides by $169$,we get $169(x^2 + y^2) = 1$.

If a circle with center $(0, 0)$ touches the line $5x + 12y = 1$,then its equation is:

Similar Questions

The equation of the circle passing through the points $(1, -2)$ and $(4, -3)$ and whose centre lies on the line $3x + 2y = 7$ is

The centre of the circle given by the parametric equations $x = 2 + 3\cos \theta$ and $y = 3\sin \theta - 1$ is

$A$ circle touches the axes at the points $(3, 0)$ and $(0, -3)$. The centre of the circle is

Circles are drawn through the point $(2,0)$ to cut intercepts of length $5$ units on the $X$-axis. If their centre lies in the first quadrant,then their equation is

$A$ circle has its equation in the form $x^2 + y^2 + 2x + 4y + 1 = 0$. Choose the correct coordinates of its centre and the right value of its radius from the following:

Vedclass Test Series

Exam Paper Generator

Online Exam Module