Find the center and radius of this circle x^2+y^2-6x-10y+30=0

The circle's centre is at the position (3, 5), and it has a radius of 2

Step-by-step explanation:

First let's put it in a useful format by completing the squares:

x² + y² - 6x - 10y + 30 = 0

x² - 6x + y² - 10y = -30

x² - 6x + 9 + y² - 10y + 25 = -30 + 9 + 25

(x - 3)² + (y - 5)² = 4

This tells us that the centre position is (3, 5) and the radius is √4, or 2

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

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step-by-step explanation: