The equation of the tangents to the circle x^ | vedclass

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(B) The given circle is $x^2 + y^2 + 4x - 4y + 4 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 2$,$f = -2$,and $c = 4$.The center is

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$x + y = 2\sqrt{2}$

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(B) The given circle is $x^2 + y^2 + 4x - 4y + 4 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 2$,$f = -2$,and $c = 4$.The center is $(-g, -f) = (-2, 2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 4 - 4} = 2$.$A$ line making equal intercepts on the positive coordinate axes has the form $x + y = a$,where $a > 0$,or $x + y - a = 0$.Since this line is a tangent to the circle,the perpendicular distance from the center $(-2, 2)$ to the line must equal the radius $r = 2$.Thus,$\left| \frac{-2 + 2 - a}{\sqrt{1^2 + 1^2}} \right| = 2$.$\left| \frac{-a}{\sqrt{2}} \right| = 2 \Rightarrow |-a| = 2\sqrt{2}$.Since the intercepts are positive,$a = 2\sqrt{2}$.Therefore,the equation of the tangent is $x + y = 2\sqrt{2}$.

The equation of the tangents to the circle $x^2 + y^2 + 4x - 4y + 4 = 0$ which make equal intercepts on the positive coordinate axes is given by

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