### Component 1
<description>
Visualizing the given parabolas and identifying their vertices.
</description>

<plan>
Okay, so I need to find the radius squared of the circle that passes through all four intersection points of the two parabolas given: y = (x + 1)^2 and x + 4 = (y - 3)^2. Hmm, let me start by trying to visualize these parabolas and their intersections.
</plan>

<prompt>
Visualize the given parabolas and identify their vertices.
</prompt>

<execution>
First, the parabola y = (x + 1)^2 is a standard upward-opening parabola shifted 1 unit to the left. Its vertex is at (-1, 0). The other parabola is given by x + 4 = (y - 3)^2. Let me rearrange this equation to make it more familiar. If I solve for x, it becomes x = (y - 3)^2 - 4. That's a right-opening parabola with vertex at (-4, 3). So this parabola opens to the right, and its vertex is at (-4, 3).
</execution>


### Component 2
<description>
Planning the strategy to find the radius squared of the circle passing through the intersection points of the parabolas.
</description>

<plan>
Now, these two parabolas should intersect at four points. The problem states that all four points lie on a circle, and we need to find the square of the radius of that circle. Interesting. So, the strategy here might be to find the four intersection points and then determine the equation of the circle passing through them.
</plan>

<prompt>
Plan the strategy to find the radius squared of the circle passing through the intersection points of the parabolas.
</prompt>

<execution>
Once we have the circle's equation, we can compute the radius squared.

But wait, solving for the intersection points might involve solving a system of equations that could result in a quartic equation. That might be complicated. Maybe there's a smarter way to use the properties of the parabolas and circles to find the radius without having to compute all four points explicitly. Let me think.
</execution>


### Component 3
<description>
Substituting one parabola's equation into the other to find the intersection points.
</description>

<plan>
First, let's try to find the equations of the intersection points. So, we have:

1. y = (x + 1)^2
2. x + 4 = (y - 3)^2

If I substitute equation 1 into equation 2, replacing y with (x + 1)^2 in equation 2, we get:
</plan>

<prompt>
Substitute one parabola's equation into the other to find the intersection points.
</prompt>

<execution>
x + 4 = [(x + 1)^2 - 3]^2

Let me simplify the right-hand side. Let's first compute (x + 1)^2 - 3:

(x + 1)^2 = x^2 + 2x + 1

So, subtracting 3 gives x^2 + 2x + 1 - 3 = x^2 + 2x - 2

Therefore, the equation becomes:

x + 4 = (x^2 + 2x - 2)^2
</execution>


### Component 4
<description>
Expanding and simplifying the equation to find the intersection points.
</description>

<plan>
Expanding the right-hand side:

(x^2 + 2x - 2)^2 = x^4 + 4x^3 + (4x^2 - 4x^2) + ... Wait, perhaps I should do this step by step.
</plan>

<prompt>
Expand and simplify the equation to find the intersection points.
</prompt>

<execution>
Let me denote A = x^2 + 2x - 2. Then A^2 = (x^2 + 2x - 2)(x^2 + 2x - 2)

Multiply term by term:

First, x^2 * x^2 = x^4

x^2 * 2x = 2x^3

x^2 * (-2) = -2x^2

2x * x^2 = 2x^3

2x * 2x = 4x^2

2x * (-2) = -4x

(-2) * x^2 = -2x^2

(-2) * 2x = -4x

(-2) * (-2) = 4

Now, combine like terms:

x^4

2x^3 + 2x^3 = 4x^3

-2x^2 + 4x^2 - 2x^2 = 0x^2

-4x -4x = -8x

+4

So, the expansion is x^4 + 4x^3 -8x +4.
</execution>


### Component 5
<description>
Rearranging and factoring the equation to find the solutions for x.
</description>

<plan>
Therefore, the equation becomes:

x + 4 = x^4 + 4x^3 -8x +4

Let's bring all terms to one side:
</plan>

<prompt>
Rearrange and factor the equation to find the solutions for x.
</prompt>

<execution>
x^4 + 4x^3 -8x +4 - x -4 = 0

Simplify:

x^4 + 4x^3 -9x = 0

Factor:

x(x^3 + 4x^2 -9) = 0
</execution>


### Component 6
<description>
Solving the cubic equation to find the remaining solutions for x.
</description>

<plan>
So, one solution is x = 0. The other solutions come from the cubic equation x^3 + 4x^2 -9 = 0.

Hmm, solving a cubic. That might be tricky, but maybe we can factor it. Let's try possible rational roots using the Rational Root Theorem.
</plan>

<prompt>
Solve the cubic equation to find the remaining solutions for x.
</prompt>

<execution>
The possible roots are ±1, ±3, ±9. Let's test x=1: 1 +4 -9 = -4 ≠0. x=3: 27 + 36 -9=54≠0. x=-1: -1 +4 -9= -6≠0. x=-3: -27 + 36 -9=0. Oh, x=-3 is a root!

So, factor x^3 +4x^2 -9 = (x + 3)(x^2 + ax + b). Let me perform polynomial division or use synthetic division.

Using synthetic division for root x=-3:

Coefficients: 1 (x^3), 4 (x^2), 0 (x), -9 (constant)

Bring down 1.

Multiply by -3: 1*(-3)=-3. Add to next coefficient:4 + (-3)=1.

Multiply by -3:1*(-3)=-3. Add to next coefficient:0 + (-3)=-3.

Multiply by -3: (-3)*(-3)=9. Add to last coefficient: -9 +9=0. Perfect.

So, the cubic factors as (x +3)(x^2 +x -3). Therefore, the quartic equation factors as x(x +3)(x^2 +x -3)=0.

Therefore, the solutions are x=0, x=-3, and the roots of x^2 +x -3=0, which are x = [-1 ± sqrt(1 +12)]/2 = [-1 ± sqrt(13)]/2.

So, four solutions for x: 0, -3, [-1 + sqrt(13)]/2, [-1 - sqrt(13)]/2.
</execution>


### Component 7
<description>
Calculating the corresponding y-values for each x-value to find the intersection points.
</description>

<plan>
Now, for each x, we can find the corresponding y using equation 1: y = (x +1)^2.

Let's compute the y-values:
</plan>

<prompt>
Calculate the corresponding y-values for each x-value to find the intersection points.
</prompt>

<execution>
1. x=0: y=(0+1)^2=1. So point (0,1).

2. x=-3: y=(-3 +1)^2=(-2)^2=4. So point (-3,4).

3. x = [-1 + sqrt(13)]/2: y = ( [-1 + sqrt(13)]/2 +1 )^2 = ( [1 + sqrt(13)]/2 )^2 = (1 + 2 sqrt(13) +13)/4 = (14 + 2 sqrt(13))/4 = (7 + sqrt(13))/2.

So, point ( [-1 + sqrt(13)]/2 , (7 + sqrt(13))/2 ).

4. x = [-1 - sqrt(13)]/2: y = ( [-1 - sqrt(13)]/2 +1 )^2 = ( [1 - sqrt(13)]/2 )^2 = (1 - 2 sqrt(13) +13)/4 = (14 - 2 sqrt(13))/4 = (7 - sqrt(13))/2.

So, point ( [-1 - sqrt(13)]/2 , (7 - sqrt(13))/2 ).

So, the four intersection points are:

(0,1), (-3,4), ( [-1 + sqrt(13)]/2 , (7 + sqrt(13))/2 ), and ( [-1 - sqrt(13)]/2 , (7 - sqrt(13))/2 ).
</execution>


### Component 8
<description>
Planning to find the circle passing through the intersection points.
</description>

<plan>
Now, we need to find the circle passing through these four points. Once we have the equation of the circle, we can compute the radius squared.

But solving for the circle equation through four points might be tedious. Let me recall that three non-colinear points determine a circle, so if these four points lie on a circle, then the fourth point must satisfy the circle equation defined by the other three. But maybe there's a better approach.
</plan>

<prompt>
Plan to find the circle passing through the intersection points.
</prompt>

<execution>
Alternatively, since all four points lie on a circle, their combined equations must satisfy the general equation of a circle: x^2 + y^2 + D x + E y + F = 0. The radius squared is (D/2)^2 + (E/2)^2 - F. So, if we can find D, E, F such that all four points satisfy the equation, then compute r^2 from that.

But to find D, E, F, we can set up a system of equations using the four points. However, since four points lie on the same circle, the system should be consistent, and we can use three of them to solve for D, E, F and verify with the fourth.
</execution>


### Component 9
<description>
Setting up and solving the system of equations to find the circle's equation.
</description>

<plan>
Let's proceed step by step.

First, let's note the four points:

1. (0,1)
2. (-3,4)
3. (a, b), where a = [-1 + sqrt(13)]/2, b = (7 + sqrt(13))/2
4. (c, d), where c = [-1 - sqrt(13)]/2, d = (7 - sqrt(13))/2

Let me label points 3 and 4 as (a, b) and (c, d) for simplicity.
</plan>

<prompt>
Set up and solve the system of equations to find the circle's equation.
</prompt>

<execution>
So, plugging each point into the general circle equation x^2 + y^2 + D x + E y + F = 0:

For point (0,1):

0^2 + 1^2 + D*0 + E*1 + F = 0 => 1 + E + F = 0 => E + F = -1. (Equation 1)

For point (-3,4):

(-3)^2 + 4^2 + D*(-3) + E*4 + F = 0 => 9 + 16 -3D +4E + F = 0 => 25 -3D +4E + F = 0. (Equation 2)

For point (a, b):

a^2 + b^2 + D*a + E*b + F = 0. (Equation 3)

For point (c, d):

c^2 + d^2 + D*c + E*d + F = 0. (Equation 4)

But since points 3 and 4 are symmetric (they are conjugates with sqrt(13) terms), maybe equations 3 and 4 are related or can be combined in a way that simplifies the system. Let's explore that.

First, compute Equations 1 and 2:

From Equation 1: E + F = -1.

From Equation 2: 25 -3D +4E + F = 0.

But since F = -1 - E (from Equation 1), substitute into Equation 2:

25 -3D +4E + (-1 - E) = 0 => 25 -3D +4E -1 -E =0 =>24 -3D +3E=0 => -3D +3E= -24 => -D + E = -8 => E = D -8.

So, now, we have E = D -8, and F = -1 - E = -1 - (D -8) = -1 - D +8 = 7 - D.

Therefore, now, Equations 1 and 2 have given us E and F in terms of D. Now, we can use Equation 3 (and 4) to solve for D.
</execution>


### Component 10
<description>
Simplifying the system of equations using symmetry properties.
</description>

<plan>
But Equation 3 is:

a^2 + b^2 + D*a + E*b + F = 0

Substituting E = D -8, F =7 - D:

a^2 + b^2 + D*a + (D -8)*b +7 - D =0

Similarly for Equation 4:

c^2 + d^2 + D*c + (D -8)*d +7 - D =0

But given the symmetry between points 3 and 4 (since they are conjugate terms), perhaps adding Equations 3 and 4 would eliminate the sqrt(13) terms, making it easier to handle.
</plan>

<prompt>
Simplify the system of equations using symmetry properties.
</prompt>

<execution>
None
</execution>


### Component 11
<description>
Solving for D, E, and F using the simplified system of equations.
</description>

<plan>
Let me compute a + c and a*c, etc. , to find expressions for a^2 + c^2, b^2 + d^2, etc.

First, let's note that:
</plan>

<prompt>
Solve for D, E, and F using the simplified system of equations.
</prompt>

<execution>
a = [-1 + sqrt(13)]/2

c = [-1 - sqrt(13)]/2

So, a + c = [ -1 + sqrt(13) -1 - sqrt(13) ] /2 = (-2)/2 = -1

a*c = [ (-1)^2 - (sqrt(13))^2 ] / (2*2) = (1 -13)/4 = (-12)/4 = -3

Similarly, b = (7 + sqrt(13))/2

d = (7 - sqrt(13))/2

Thus, b + d = [7 + sqrt(13) +7 - sqrt(13)]/2 =14/2=7

b*d = [7^2 - (sqrt(13))^2]/4 = (49 -13)/4=36/4=9

Also, a^2 + c^2 = (a + c)^2 - 2ac = (-1)^2 -2*(-3) =1 +6=7

Similarly, b^2 + d^2 = (b + d)^2 -2bd =7^2 -2*9=49 -18=31

Now, let's compute Equations 3 + 4:

(a^2 + c^2) + (b^2 + d^2) + D(a + c) + (D -8)(b + d) + 2*(7 - D) =0

From above:

a^2 + c^2 =7

b^2 + d^2=31

a + c = -1

b + d=7

Thus:

7 +31 + D*(-1) + (D -8)*7 +14 -2D =0

Compute each term:

7 +31=38

-D

(D -8)*7=7D -56

14 -2D

Combine all terms:

38 - D +7D -56 +14 -2D=0

Combine like terms:

38 -56 +14= (38 +14) -56=52 -56=-4

-D +7D -2D=4D

So total equation: -4 +4D=0 =>4D=4 =>D=1.
</execution>


### Component 12
<description>
Completing the square to find the standard form of the circle's equation and the radius squared.
</description>

<plan>
So D=1. Then E=D -8=1 -8=-7. F=7 -D=7 -1=6.

Therefore, the equation of the circle is x^2 + y^2 +1x -7y +6=0.

We can write this in standard form by completing the squares.
</plan>

<prompt>
Complete the square to find the standard form of the circle's equation and the radius squared.
</prompt>

<execution>
Starting with x^2 + x + y^2 -7y = -6

Complete the square for x:

x^2 +x = (x + 0.5)^2 -0.25

Complete the square for y:

y^2 -7y = (y - 3.5)^2 - (3.5)^2 = (y - 7/2)^2 - (49/4)

So substituting back:

(x + 0.5)^2 -0.25 + (y - 7/2)^2 -49/4 = -6

Combine constants:

-0.25 -49/4 = (-1/4 -49/4)= -50/4= -25/2

Thus:

(x + 0.5)^2 + (y - 7/2)^2 -25/2 = -6

Bring -25/2 to the right:

(x + 0.5)^2 + (y - 7/2)^2 = -6 +25/2 = (-12/2 +25/2)=13/2

Therefore, the radius squared is 13/2, so r^2=13/2.
</execution>


### Component 13
<description>
Verifying that all four points lie on the circle.
</description>

<plan>
But wait, let me verify this result. Let's check if all four points lie on this circle.
</plan>

<prompt>
Verify that all four points lie on the circle.
</prompt>

<execution>
First, check point (0,1):

(x +0.5)^2 + (y -7/2)^2 = (0.5)^2 + (1 -3.5)^2 =0.25 + (-2.5)^2=0.25 +6.25=6.5=13/2. Correct.

Point (-3,4):

(x +0.5)^2 + (y -7/2)^2= (-3 +0.5)^2 + (4 -3.5)^2= (-2.5)^2 + (0.5)^2=6.25 +0.25=6.5=13/2. Correct.

Point (a, b):

a = [-1 + sqrt(13)]/2, b=(7 + sqrt(13))/2

Compute (a +0.5)^2 + (b -7/2)^2:

a +0.5 = [ -1 + sqrt(13) ]/2 +0.5 = [ -1 + sqrt(13) +1 ]/2 = sqrt(13)/2

So (a +0.5)^2 = (sqrt(13)/2)^2 =13/4

b -7/2= [ (7 + sqrt(13))/2 -7/2 ]= sqrt(13)/2

Thus, (b -7/2)^2 = (sqrt(13)/2)^2=13/4

Sum:13/4 +13/4=26/4=13/2. Correct.

Similarly for point (c,d):

c = [-1 - sqrt(13)]/2, d=(7 - sqrt(13))/2

(c +0.5)^2 = [ -1 - sqrt(13) ]/2 +0.5= [ -1 - sqrt(13) +1 ]/2= -sqrt(13)/2, squared is13/4

(d -7/2)^2= [ (7 - sqrt(13))/2 -7/2 ]= -sqrt(13)/2, squared is13/4

Sum:13/4 +13/4=13/2. Correct.
</execution>


### Component 14
<description>
Exploring an alternative approach to confirm the result.
</description>

<plan>
Therefore, the answer is 13/2.

But just to make sure, let me check if there's an alternative approach that might lead to the same result without solving for all the points. For instance, since the two parabolas intersect at four points lying on a circle, maybe the equation of the circle can be derived by combining the equations of the parabolas in a certain way.
</plan>

<prompt>
Explore an alternative approach to confirm the result.
</prompt>

<execution>
Given the two equations:

1. y = (x +1)^2

2. x +4 = (y -3)^2

If we can find a linear combination of these equations that gives the equation of a circle, that might work. Let's try to manipulate them.

From equation 1: y = x^2 + 2x +1. Let's rearrange this as x^2 +2x +1 - y =0.

From equation 2: x +4 = (y -3)^2. Let's expand the right-hand side: (y -3)^2 = y^2 -6y +9. So, x +4 = y^2 -6y +9. Rearranged: y^2 -6y +9 -x -4=0 => y^2 -6y -x +5=0.

So now we have:

Equation A: x^2 +2x +1 - y =0

Equation B: y^2 -6y -x +5=0

If we can find constants α and β such that α*(Equation A) + β*(Equation B) =0 gives the equation of a circle. Since a circle equation has x^2 and y^2 terms with the same coefficient (assuming no xy term). Let's try to eliminate some terms.

Multiply Equation A by some α and Equation B by some β and add them:

α(x^2 +2x +1 - y) + β(y^2 -6y -x +5) =0

Expanding:

αx^2 +2αx +α -αy + βy^2 -6βy -βx +5β =0

Combine like terms:

x² term: αx²

y² term: βy²

x term: (2α -β)x

y term: (-α -6β)y

constants: α +5β

To make this a circle equation, we need the coefficients of x² and y² to be equal. So set α=β.

Let’s set α=β=1 (for simplicity):

Then:

x² + y² + (2*1 -1)x + (-1 -6*1)y + (1 +5*1) =0

Simplify:

x² + y² +1x -7y +6=0

Which is exactly the circle equation we found earlier! So, indeed, adding Equation A and Equation B gives the circle equation. Therefore, the intersection points lie on this circle. Therefore, this is the circle we are looking for, and its radius squared is 13/2.

So, I think this confirms the answer.
</execution>


### Component 15
<description>
Presenting the final answer.
</description>

<plan>
None
</plan>

<prompt>
Present the final answer.
</prompt>

<execution>
**Final Answer**
The value of \( r^2 \) is \(\boxed{\dfrac{13}{2}}\).
</execution>
