Find the center and radius of the circle whose equation is x2 + y2 - 6x-2y+4 = 0

The center of the circle = (3, 1)

The radius of a circle is given as r = √6

Step-by-step explanation:

The equation of a circle is given as:

(x - a) ² + (y - b) ² = r²

where (a, b) = center of the circle

r = radius of the the circle

In the question above, we are given the equation of a circle as

x² + y² - 6x - 2y + 4 = 0

In order to find the center and radius of this circle we would use completing the square method to solve it

Step 1

Collect the like terms

x² - 6x + y² - 2y + 4 = 0

Step 2

Complete the square for both x and y

x² - 6x + (-6/2) ² + y² - 2y + (-2/2) ² = - 4 + ( - 6/2) ² + (-2/2) ²

x² - 6x + (-3) ² + y² - 2y + (-1) ² = - 4 + (3) ² + (1) ²

x² - 6x + 9 + y² - 2y + 1 = - 4 + 9 + 1

(x² - 6x + 9) + (y² - 2y + 1) = - 4 + 9 + 1

(x² - 3x - 3x + 9) + (y² - y - y + 1) = 6

x (x - 3) - 3 (x - 3) + y (y - 1) - 1 (y - 1) = 6

(x - 3) (x - 3) + (y - 1) (y - 1) = 6

(x - 3) ² (y - 1) ² = 6

Since,

The equation of a circle is given as:

(x - a) ² + (y - b) ² = r²

where (a, b) = center of the circle

r = radius of the the circle

The equation of the the circle for the above question is calculated as:

(x - 3) ² (y - 1) ² = 6

where (a, b) = center of the circle = (3, 1)

r = radius of the the circle

r² = 6

r = √6