The equation of a circle whose centre is the | vedclass

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(C) The distance between the lines $x = 1$ and $x = -1$ is $|1 - (-1)| = |1 + 1| = 2$.Thus,the radius of the circle is $r = 2$.The centre of the circl

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$x^2 + y^2 = 4$

Vedclass · Answer

(C) The distance between the lines $x = 1$ and $x = -1$ is $|1 - (-1)| = |1 + 1| = 2$.Thus,the radius of the circle is $r = 2$.The centre of the circle is the origin $(0, 0)$.The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - 0)^2 + (y - 0)^2 = 2^2$.Therefore,the equation of the circle is $x^2 + y^2 = 4$.

The equation of a circle whose centre is the origin and whose radius is equal to the distance between the lines $x = 1$ and $x = -1$ is:

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