How can I find the radius of a circle given the equation x² + y² – 8x – 6y – 21 = 0?

To find the radius of a circle from its equation, we first need to rewrite the given equation in the standard form of a circle’s equation. The standard form is:

(x – h)² + (y – k)² = r²

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

Here, you have the equation:

x² + y² – 8x – 6y – 21 = 0

We will start by reorganizing this equation:

x² – 8x + y² – 6y = 21

Next, we complete the square for both the x and y terms.

Completing the square for x:

Take the coefficient of x, which is -8, divide it by 2 to get -4, and then square it to get 16.

Add and subtract 16:

(x² – 8x + 16) – 16

Completing the square for y:

Take the coefficient of y, which is -6, divide it by 2 to get -3, and then square it to get 9.

Add and subtract 9:

(y² – 6y + 9) – 9

Now substituting these completed squares back into the equation gives us:

(x – 4)² – 16 + (y – 3)² – 9 = 21

Simplifying this, we combine constants:

(x – 4)² + (y – 3)² – 25 = 21

This leads to:

(x – 4)² + (y – 3)² = 46

Now we can see that this is in the standard form of a circle’s equation, where:

  • The center of the circle is at (4, 3)
  • The right side of the equation, 46, represents r².

To find the radius, take the square root of 46:

r = √46

Thus, the radius of the circle is:

r ≈ 6.78

This means the radius of the circle described by the equation x² + y² – 8x – 6y – 21 = 0 is approximately 6.78 units.

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