Question 298099
Rewrite in standard form:
{{{x^2+y^2-4x+8y+4 = 0}}} Group the x-terms and the y-terms as shown:
{{{(x^2-4x)+(y^2+8y)+4 = 0}}} Subtract 4 from both sides.
{{{x^2-4x)+(y^2+8y) = -4}}} Now complete the square in x and y by adding the square of half the x-term coefficient and the square of half the y-term coefficient to both sides of the equation.
{{{(x^2-4x+highlight((-4/2)^2))+(y^2+8y+highlight_green((8/2)^2)) = -4+highlight((-4/2)^2)+highlight_green((8/2)^2)}}} Simplify both sides of the equation.
{{{(x^2-4x+4)+(y^2+8y+16) = -4+4+16}}} Now factor the left side and simplify the right side.
{{{(x-2)^2+(y+4)^2 = 16}}} Finally...
{{{highlight((x-2)^2+(y+4)^2 = 4^2)}}} Compare with standard form...
{{{(x+h)^2+(y+k)^2 = r^2}}} and you can see that {{{h = -2}}},{{{k = 4}}}, and {{{r = 4}}} and this represents a circle with center at (-2, 4) and radius of 4.