Question

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

Answer 1

Answer:

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