What is the radius of the circle x^2+y^2-4y=21 7 Write your answer in simplified, rationalized form. square

To find the radius of the circle given by the equation \(x^2 + y^2 - 4y = 21\), we need to rewrite the equation in the standard form of a circle's equation, which is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.<br /><br />First, let's complete the square for the \(y\)-terms in the given equation:<br /><br />\[ x^2 + y^2 - 4y = 21 \]<br /><br />We focus on the \(y\)-terms: \(y^2 - 4y\). To complete the square, we add and subtract the square of half the coefficient of \(y\):<br /><br />\[ y^2 - 4y \]<br />\[ = y^2 - 4y + 4 - 4 \]<br />\[ = (y - 2)^2 - 4 \]<br /><br />Now substitute this back into the original equation:<br /><br />\[ x^2 + (y - 2)^2 - 4 = 21 \]<br /><br />Add 4 to both sides to isolate the completed square term:<br /><br />\[ x^2 + (y - 2)^2 = 25 \]<br /><br />This is now in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \(h = 0\), \(k = 2\), and \(r^2 = 25\). Therefore, the radius \(r\) is:<br /><br />\[ r = \sqrt{25} = 5 \]<br /><br />So, the radius of the circle is \(\boxed{5}\).