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12 May, 01:53

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

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  1. 12 May, 02:02
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    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
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