To solve the problem, we need to find the points on the circle that have tangents parallel to the line joining the origin to the center of the circle.
### Step 1: Write the general equation of the circle
The general equation of a circle can be written as:
\[
x^2 + y^2 + 2gx + 2fy + c = 0
\]
### Step 2: Substitute the points into the equation
The circle passes through the points (-1, 1), (0, 6), and (5, 5). We will substitute these points into the circle's equation to form equations for \(g\), \(f\), and \(c\).
1. For the point (-1, 1):
\[
(-1)^2 + (1)^2 + 2g(-1) + 2f(1) + c = 0
\]
This simplifies to:
\[
1 + 1 - 2g + 2f + c = 0 \quad \Rightarrow \quad -2g + 2f + c = -2 \quad \text{(Equation 1)}
\]
2. For the point (0, 6):
\[
(0)^2 + (6)^2 + 2g(0) + 2f(6) + c = 0
\]
This simplifies to:
\[
36 + 12f + c = 0 \quad \Rightarrow \quad 12f + c = -36 \quad \text{(Equation 2)}
\]
3. For the point (5, 5):
\[
(5)^2 + (5)^2 + 2g(5) + 2f(5) + c = 0
\]
This simplifies to:
\[
25 + 25 + 10g + 10f + c = 0 \quad \Rightarrow \quad 10g + 10f + c = -50 \quad \text{(Equation 3)}
\]
### Step 3: Solve the system of equations
Now we have three equations:
1. \(-2g + 2f + c = -2\) (Equation 1)
2. \(12f + c = -36\) (Equation 2)
3. \(10g + 10f + c = -50\) (Equation 3)
From Equation 1, we can express \(c\) in terms of \(g\) and \(f\):
\[
c = -2 + 2g - 2f \quad \text{(Substituting into Equation 2)}
\]
Substituting into Equation 2:
\[
12f + (-2 + 2g - 2f) = -36
\]
This simplifies to:
\[
10f + 2g = -34 \quad \Rightarrow \quad 5f + g = -17 \quad \text{(Equation 4)}
\]
Now substitute \(c\) into Equation 3:
\[
10g + 10f + (-2 + 2g - 2f) = -50
\]
This simplifies to:
\[
12g + 8f = -48 \quad \Rightarrow \quad 3g + 2f = -12 \quad \text{(Equation 5)}
\]
### Step 4: Solve Equations 4 and 5
Now we have:
1. \(5f + g = -17\) (Equation 4)
2. \(3g + 2f = -12\) (Equation 5)
From Equation 4, we can express \(g\) in terms of \(f\):
\[
g = -17 - 5f
\]
Substituting into Equation 5:
\[
3(-17 - 5f) + 2f = -12
\]
This simplifies to:
\[
-51 - 15f + 2f = -12 \quad \Rightarrow \quad -13f = 39 \quad \Rightarrow \quad f = -3
\]
Now substituting \(f = -3\) back into Equation 4:
\[
5(-3) + g = -17 \quad \Rightarrow \quad -15 + g = -17 \quad \Rightarrow \quad g = -2
\]
### Step 5: Find \(c\)
Substituting \(g\) and \(f\) back into Equation 1 to find \(c\):
\[
-2(-2) + 2(-3) + c = -2 \quad \Rightarrow \quad 4 - 6 + c = -2 \quad \Rightarrow \quad c = 0
\]
### Step 6: Write the equation of the circle
Now we have \(g = -2\), \(f = -3\), and \(c = 0\). The equation of the circle is:
\[
x^2 + y^2 - 4x - 6y = 0
\]
### Step 7: Find the center of the circle
The center of the circle is given by \((-g, -f)\):
\[
(2, 3)
\]
### Step 8: Find the slope of the line joining the center to the origin
The slope of the line joining the origin (0, 0) to the center (2, 3) is:
\[
\text{slope} = \frac{3 - 0}{2 - 0} = \frac{3}{2}
\]
### Step 9: Find the equation of the tangent line
The slope of the tangent line that is parallel to this line is also \(\frac{3}{2}\). The equation of the tangent line at point \((x_0, y_0)\) on the circle can be written as:
\[
y - y_0 = \frac{3}{2}(x - x_0)
\]
### Step 10: Substitute into the circle's equation
We need to find points \((x_0, y_0)\) on the circle such that the tangent line is parallel to the line joining the origin to the center. Substitute \(y = \frac{3}{2}x + b\) into the circle's equation and solve for \(x\) and \(y\).
After solving, we find the points of tangency to be:
1. \((-1, 5)\)
2. \((5, 1)\)
### Final Answer
The points on the circle, the tangents at which are parallel to the straight line joining the origin to its center, are:
\[
(-1, 5) \quad \text{and} \quad (5, 1)
\]
