Find in each case whether the given point lies inside, outside or on the given circle. (−4, 1) , x^2 + y^2 + 2x + 8y − 13 = 0

To determine whether the given point \((-4, 1)\) lies inside, outside, or on the circle defined by the equation \(x^2 + y^2 + 2x + 8y - 13 = 0\), we need to compare the point's distance from the circle's center with the circle's radius.

### Step 1: Rewrite the Circle's Equation in Standard Form

The equation of the circle is given as:
\[
x^2 + y^2 + 2x + 8y - 13 = 0
\]

We will complete the square for both \(x\) and \(y\) terms:

1. For the \(x\) terms: \(x^2 + 2x\)
   - Add and subtract \(\left(\frac{2}{2}\right)^2 = 1\), so it becomes \((x + 1)^2 - 1\).

2. For the \(y\) terms: \(y^2 + 8y\)
   - Add and subtract \(\left(\frac{8}{2}\right)^2 = 16\), so it becomes \((y + 4)^2 - 16\).

Substituting these back into the equation:
\[
(x + 1)^2 - 1 + (y + 4)^2 - 16 - 13 = 0
\]

Simplify the equation:
\[
(x + 1)^2 + (y + 4)^2 - 30 = 0

Learn how to determine if the point (-4, 1) lies inside, outside, or on the circle defined by the equation x^2 + y^2 + 2x + 8y - 13 = 0. This problem involves circle equations and completing the square technique in algebra. Suitable for high school students.

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