PRE-CALCULUS

Hello!

First of all, let's evaluate the center (C) and radius (R) of the given equation

• x² + y² - 4x + 8y - 5 = 0

We need to rewrite the equation in standard form, which is:

• (x - h)² + (y - k)² = r²

Where (h, k) is the center of the circle, and r is the radius.

To do this, we complete the square for both x and y terms:

• (x² - 4x) + (y² + 8y) = 5

• (x² - 4x) + (y² + 8y) = 5(x² - 4x + 4) + (y² + 8y + 16) = 5 + 4 + 16

• (x² - 4x) + (y² + 8y) = 5(x² - 4x + 4) + (y² + 8y + 16) = 5 + 4 + 16(x - 2)² + (y + 4)² = 25

Now we can see that the equation is in standard form, with center (h, k) = (2, -4) and radius r = √25 = 5.

Answer➜The center of the circle is at (2, -4) and its radius is 5.

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Verified answer

PRE-CALCULUS

Evaluate the center and radius of this equation:

• [tex]\large\rm{x^2 + y^2 - 4x + 8y - 5 = 0}[/tex]

Answer:

The center is (2,-4) and the Radius is 3.

Solution and Explanation:

To find the center and radius of the circle with equation:

• [tex]\large\rm{x^2 + y^2 - 4x + 8y - 5 = 0}[/tex]

We need to complete the square for both x and y terms. We start by rearranging the equation:

• [tex]\large\rm{x^2 - 4x + y^2 + 8y = 5}[/tex]

To complete the square for the x terms, we add and subtract (4/2)² = 4 inside the parentheses:

• [tex]\large\rm{(x^2 - 4x + 4) + y^2 + 8y = 9}[/tex]

Simplify the expression:

• [tex]\large\rm{(x - 2)^2 + (y + 4)^2 = 3^2}[/tex]

Comparing this with the standard form of the equation of a circle:

• [tex]\large\rm{(x - h)^2 + (y - k)^2 = r^2}[/tex]

We see that the center of the circle is (2, -4) and the radius is 3.

Therefore,

• Center = (2,-4)
• Radius = 3

The center is obtained by completing the square for both x and y, while the radius is the square root of the constant term in the completed square form.

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