State whether the following are true or false
(d) the (-1, 5) lies inside the circle` x^2+y^2-2x+6y+1`

Answer

To determine whether the point (-1, 5) lies inside the circle defined by the equation \( x^2 + y^2 - 2x + 6y + 1 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 - 2x + 6y + 1 = 0 \]
We can rearrange this equation to find the center and radius of the circle.

### Step 2: Complete the Square
To rewrite the equation in standard form, we complete the square for the \(x\) and \(y\) terms.

For \(x\):
\[ x^2 - 2x \quad \text{can be rewritten as} \quad (x - 1)^2 - 1 \]

For \(y\):
\[ y^2 + 6y \quad \text{can be rewritten as} \quad (y + 3)^2 - 9 \]

Now substituting back into the equation:
\[ (x - 1)^2 - 1 + (y + 3)^2 - 9 + 1 = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y + 3)^2 - 9 = 0 \]
or
\[ (x - 1)^2 + (y + 3)^2 = 9 \]

### Step 3: Identify the Center and Radius
From the equation \((x - 1)^2 + (y + 3)^2 = 9\), we can see that:
- The center of the circle is at \((1, -3)\)
- The radius of the circle is \(r = \sqrt{9} = 3\)

### Step 4: Calculate the Distance from the Center to the Point
Next, we calculate the distance from the center \((1, -3)\) to the point \((-1, 5)\) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{((-1) - 1)^2 + (5 - (-3))^2} \]
\[ = \sqrt{(-2)^2 + (5 + 3)^2} \]
\[ = \sqrt{4 + 64} \]
\[ = \sqrt{68} = 2\sqrt{17} \]

### Step 5: Compare the Distance with the Radius
Now we compare the distance \(d\) with the radius \(r\):
- The radius \(r = 3\)
- The distance \(d = 2\sqrt{17}\)

Calculating \(2\sqrt{17}\):
\[ \sqrt{17} \approx 4.123 \quad \Rightarrow \quad 2\sqrt{17} \approx 8.246 \]
Since \(8.246 > 3\), the point \((-1, 5)\) lies outside the circle.

### Conclusion
Since the point \((-1, 5)\) does not lie inside the circle, the statement is **False**.

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