find the center and radius of a circle with the given equation x^2+y ^2-10x+10y=0

x^2+y ^2-10x+10y=0 The goal is to write this equation in the form (x - h)^2 + (y - k) ^2 = r^2 which is standard form for the equation of a circle.

Compare the equation of your circle with\( \quad ax^2+by^2+2gx+2hy+c=0 \) then centre of this circle will be \(C = (-g,-h)\) and radius will be \(r=\sqrt{g^2+h^2-c}\).

x^2 -10x + ____ + y ^2 + 10y + ___ = 0 + ___ + ____

To get to the standard form, we have to complete the square.

x^2 -10x + __25__ + y ^2 + 10y + _25__ = 0 + __25_ + __25__

To determine that 25 is to be added to complete the square, take half (-10) and square it. The same applies to the coefficient of the linear y term. If 25 is added on one side of the equation, it must be added to the other side to maintain equality.

x^2 -10x + __25__ + y ^2 + 10y + _25__ = 0 + __25_ + __25__ (x + 5) ^2 + (y + 5)^2 = 50

(x + 5) ^2 + (y + 5)^2 = 50 (x + 5)² + (y + 5)² = (5√2)²

This is the equation of a circle in standard form. The center of the circle is (-5,-5) and the radius is (5√2).

Graph courtesy of "The Wolf" at this link: http://tinyurl.com/c9dbu5n

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