<Question>
For what value of $c$ will the circle with equation $x^2 - 10x + y^2 + 6y + c = 0$ have a radius of length 1?
</Question>

<think>
<Plan_1>
Okay, so I need to find the value of c such that the circle given by the equation x² - 10x + y² + 6y + c = 0 has a radius of 1. Hmm, let me think. I remember that the general equation of a circle is something like (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. The given equation isn't in that standard form, so I think I need to complete the square for both the x and y terms to rewrite it in the standard form.

Let me start with the x terms.
The equation has x² - 10x. To complete the square, I need to take half of the coefficient of x, which is -10. Half of -10 is -5, and then squaring that gives (-5)² = 25. So, I can write x² - 10x as (x - 5)² - 25. Because (x - 5)² expands to x² - 10x + 25, so to get back to the original expression, we subtract 25.

Now for the y terms: y² + 6y.
Similarly, completing the square here. Take half of 6, which is 3, and squaring it gives 9. So, y² + 6y can be written as (y + 3)² - 9. Because (y + 3)² is y² + 6y + 9, so subtracting 9 gives the original expression.

So substituting these back into the original equation:

<prompt_1.1> Substitute the completed square forms back into the original equation and simplify. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
(x - 5)² - 25 + (y + 3)² - 9 + c = 0

Combine the constants: -25 -9 + c = -34 + c. So the equation becomes:

(x - 5)² + (y + 3)² + (c - 34) = 0

Which is the same as:

(x - 5)² + (y + 3)² = 34 - c
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But we want the radius to be 1. In the standard form, the right-hand side is r². So, 34 - c must equal 1² = 1. Therefore, solving for c:
34 - c = 1

Subtract 34 from both sides: -c = 1 - 34

Which is -c = -33. Multiply both sides by -1: c = 33.

Wait, let me check that again. If 34 - c = 1, then c = 34 - 1 = 33.

Let me verify by plugging c back into the original equation.

<prompt_2.1> Verify the solution by plugging the value of c back into the original equation. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Original equation: x² -10x + y² +6y +33=0

Completing the squares as before:

(x -5)² -25 + (y +3)² -9 +33=0

Combine constants: -25 -9 +33 = (-34) +33 = -1

So, (x -5)² + (y +3)² -1 =0 → (x -5)² + (y +3)² =1. That's a circle with radius 1.
</execution_2.1>
</Execution_2>

<Final_answer>
Therefore, the value of $c$ for which the circle has a radius of length 1 is $c = \boxed{33}$.
</Final_answer>
</think>