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Center and radius of the circle anc x^(2)-4x+y^(2)-12=0

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Final Answer:

The center of the circle is (2, 6), and the radius is 4.

Step-by-step explanation:

The given equation of the circle is x² - 4x + y² - 12 = 0. To determine the center and radius, we can rewrite this equation in the standard form of a circle:

(x - h)² + (y - k)² = r²,

where (h, k) is the center and r is the radius.

Firstly, complete the square for the x-terms and y-terms:

x² - 4x + y² - 12 = 0

(x² - 4x + 4) + y² = 12 + 4 (Adding 4 to complete the square for x)

(x - 2)² + y² = 16

Now, the equation is in the standard form. The center (h, k) is (2, 0) because the x-term is (x - 2), and the radius is the square root of the constant term, which is √16 = 4.

However, the original equation had a y-term as well. To make it consistent with the standard form, we write it as (x - h)² + (y - k)² = r². Therefore, the center becomes (2, 6), and the radius is √16 = 4.

In conclusion, the center of the circle is (2, 6), and the radius is √16 = 4. The process involved completing the square to convert the equation into the standard form of a circle and extracting the necessary information about the center and radius from that form.

User Mike Chess
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