Question

the equation of a circle is given below. identify the radius and the center. then graph the circle.
$x^{2}+y^{2}-4x - 2y - 4 = 0$
radius:
center:

Explanation:

Step1: Rewrite the equation by completing the square for x and y terms.

The given equation is \(x^{2}+y^{2}-4x - 2y-4 = 0\).
Group the x - terms and y - terms: \((x^{2}-4x)+(y^{2}-2y)=4\).
For the x - terms: \(x^{2}-4x=(x - 2)^{2}-4\) (using the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=x\), \(2ab = 4x\) so \(b = 2\)).
For the y - terms: \(y^{2}-2y=(y - 1)^{2}-1\) (using the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a = y\), \(2ab=2y\) so \(b = 1\)).
Substitute these back into the equation: \((x - 2)^{2}-4+(y - 1)^{2}-1 = 4\).

Step2: Simplify the equation to the standard form of a circle.

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
Simplify the left - hand side: \((x - 2)^{2}+(y - 1)^{2}-5 = 4\).
Then, \((x - 2)^{2}+(y - 1)^{2}=4 + 5=9\).
Since \(9 = 3^{2}\), we can compare with the standard form \((x - h)^{2}+(y - k)^{2}=r^{2}\).

Answer:

Radius: \(3\)
Center: \((2,1)\)