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

In a square $ABCD$ of side length $a$,suppose $AB$ and $AD$ are along the coordinate axes. Then,the circle that circumscribes the square is

The diameters of a circle are along $2x + y - 7 = 0$ and $x + 3y - 11 = 0$. Then,the equation of this circle,which also passes through $(5, 7)$,is

In the $xy$-plane,the segment with endpoints $(3, 8)$ and $(-5, 2)$ is the diameter of a circle. The point $(k, 10)$ lies on the circle for:

If two vertices of an equilateral triangle are $A(-a, 0)$ and $B(a, 0)$,where $a > 0$,and the third vertex $C$ lies above the $x$-axis,then the equation of the circumcircle of $\Delta ABC$ is:

The centres of the circles $x^2 + y^2 = 1$,$x^2 + y^2 + 6x - 2y = 1$ and $x^2 + y^2 - 12x + 4y = 1$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module