Question 91183
Question:


 equation of a circle with the center at (-2,2) and a radius of 4?

Answer:


The general equation  of a circle whose centre is(a,b) and radius r is given by,

{{{ (x-a)^2 + (x-b)^2 = r^2 }}}


So here the equation is given by......


{{{ (x-(-2))^2 + (x-2)^2 = 4^2 }}}


{{{ (x+2)^2 + (x-2)^2 = 16 }}}, which is the equation of the given circle...


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You can expand this expression as follows.



==> {{{ x^2 + 4x + 4 + x^2 - 4x + 4 = 16 }}}



==> {{{ x^2 + x^2 + 4x - 4x   + 4 + 4 = 16 }}}



==> {{{ 2x^2 + 8 = 16 }}}


==> {{{ 2x^2 + 8 - 16 = 0 }}}



==> {{{ 2x^2 - 8 = 0 }}}


==> {{{ x^2 - 4 = 0 }}}


==> ( x+ 2 ) ( x - 2 ) = 0, which is the simplified form the equation of the given circle.



Hope you understood.


Regards.


Praseena.