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(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 $x = 2c$.

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