Find the equation of a circle that touches th | vedclass

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

Question

(D) The circle touches the lines $x = 0$ and $x = 2c$. The distance between these two parallel lines is $2c$. Thus,the diameter of the circle is $2c$,

Vedclass · Accepted Answer

$x^2 + y^2 - 2cx \pm 2cy + c^2 = 0$

Vedclass · Answer

(D) The circle touches the lines $x = 0$ and $x = 2c$. The distance between these two parallel lines is $2c$. Thus,the diameter of the circle is $2c$,and the radius $r = c$.Since the circle touches $x = 0$ and $x = 2c$,the $x$-coordinate of the center must be $h = c$.Since the circle also touches $y = 0$,the $y$-coordinate of the center must be $k = c$ or $k = -c$.Thus,the center of the circle is $(c, c)$ or $(c, -c)$.The equation of the circle is $(x - c)^2 + (y \mp c)^2 = c^2$.Expanding this,we get $x^2 - 2cx + c^2 + y^2 \mp 2cy + c^2 = c^2$.Simplifying,we get $x^2 + y^2 - 2cx \mp 2cy + c^2 = 0$.

Find the equation of a circle that touches the lines $x = 0$,$y = 0$,and $x = 2c$.

Similar Questions

Find the centre and radius of the circle $x^{2}+y^{2}-4x-8y-45=0$.

The equation of the circle whose end points of a diameter are the centres of the circles $x^{2}+y^{2}+2x-4y+1=0$ and $x^{2}+y^{2}-8x+6y+17=0$ is

Find the equation of a circle whose center lies on the $y$-axis,has a radius of $3$,and passes through the origin.

The radius of the larger circle lying in the first quadrant and touching the line $4x + 3y - 12 = 0$ and the coordinate axes is:

When does the equation $x^2 + y^2 + 2gx + 2fy + c = 0$ represent a point circle?

Vedclass Test Series

Exam Paper Generator

Online Exam Module