Week 8 ASSIgnment Score: \( 5 / 16 \) Answered: \( 4 / 14 \) Question 5 Find the center and radius of the circle whose equation is \( x^{2}-5 x+y^{2}+4 y-12=0 \). The center of the circle is ( The radius of the circle is Question Help: Video Submit Question Jump to Answer

To find the center and radius of the circle given by the equation \( x^{2}-5x+y^{2}+4y-12=0 \), you first need to rewrite it in the standard form \((x-h)^2 + (y-k)^2 = r^2\). Start by rearranging the equation: 1. Group the \(x\) terms and the \(y\) terms: \[ x^{2} - 5x + y^{2} + 4y = 12 \] 2. Complete the square for \(x\): - Take half of -5, square it: \((-5/2)^2 = 25/4\). - Add and subtract \(25/4\): \[ x^{2} - 5x + \frac{25}{4} - \frac{25}{4} \] 3. Complete the square for \(y\): - Take half of 4, square it: \((4/2)^2 = 4\). - Add and subtract \(4\): \[ y^{2} + 4y + 4 - 4 \] Putting it all together: \[ \left(x - \frac{5}{2}\right)^{2} + \left(y + 2\right)^{2} = 12 + \frac{25}{4} + 4 \] Calculate the right side: \[ 12 + \frac{25}{4} + 4 = 12 + 6.25 + 4 = 22.25 = \frac{89}{4} \] Now we can write the equation in standard form: \[ \left(x - \frac{5}{2}\right)^{2} + \left(y + 2\right)^{2} = \frac{89}{4} \] From this, we find that the center \((h, k)\) is \(\left(\frac{5}{2}, -2\right)\) and the radius \(r\) is \(\sqrt{\frac{89}{4}} = \frac{\sqrt{89}}{2}\). So, the center of the circle is \(\left(\frac{5}{2}, -2\right)\) and the radius is \(\frac{\sqrt{89}}{2}\).