Question

Find the center and radius of the circle whose equation is 20-%206x%20-%202y%20%2B%204%20%3D%200" id="TexFormula1" title="x^{2} + y^{2} - 6x - 2y + 4 = 0" alt="x^{2} + y^{2} - 6x - 2y + 4 = 0" align="absmiddle" class="latex-formula">

Answer 1

Answer:

the center is (3,1) and the radius is √6.

Step-by-step explanation:

The center of a circle can be found using the equation $(x-h)^2 + (y-k)^2 = r^2$ and is (h,k) from it. Notice h and k are the opposite value as in the equation.

First write the equation in this form.

$x^2 - 6x + ( ?)+ y^2 - 2y + (?) + 4 = 0$

Complete the square with each variable to find what numbers should go in place of the question marks.

$(-6/2)^2 = -3^2 = 9$

$(-2/2)^2 = -1^2 = 1$

Add 1 and 9 to both sides of the equation.

$x^2 - 6x + 9 + y^2 - 2y + 1 + 4 = 1 + 9\\(x-3)^2 + (y-1)^2 + 4 = 10\\(x-3)^2 + (y-1)^2 = 6$

So the center is (3,1) and the radius is √6.