To find the equation of the circle that passes through the point \((-1, 0)\) and touches the line \(x + y - 7 = 0\) at the point \((3, 4)\), we can follow these steps:
### Step 1: Understand the Problem
We need to find the center of the circle, which we will denote as \((A, B)\). The circle must pass through the point \((-1, 0)\) and touch the line \(x + y - 7 = 0\) at the point \((3, 4)\).
### Step 2: Use the Distance Formula
Since the circle touches the line at \((3, 4)\), the distance from the center \((A, B)\) to the point \((3, 4)\) must equal the distance from the center to the point \((-1, 0)\) (the radius).
Using the distance formula:
1. Distance from \((A, B)\) to \((3, 4)\):
\[
CA = \sqrt{(A - 3)^2 + (B - 4)^2}
\]
2. Distance from \((A, B)\) to \((-1, 0)\):
\[
CB = \sqrt{(A + 1)^2 + (B - 0)^2}
\]
Setting these distances equal gives us:
\[
\sqrt{(A - 3)^2 + (B - 4)^2} = \sqrt{(A + 1)^2 + B^2}
\]
### Step 3: Square Both Sides
Squaring both sides to eliminate the square roots:
\[
(A - 3)^2 + (B - 4)^2 = (A + 1)^2 + B^2
\]
### Step 4: Expand Both Sides
Expanding both sides:
\[
(A^2 - 6A + 9) + (B^2 - 8B + 16) = (A^2 + 2A + 1) + B^2
\]
This simplifies to:
\[
A^2 - 6A + 9 + B^2 - 8B + 16 = A^2 + 2A + 1 + B^2
\]
### Step 5: Cancel Common Terms
Cancel \(A^2\) and \(B^2\) from both sides:
\[
-6A - 8B + 25 = 2A + 1
\]
### Step 6: Rearrange the Equation
Rearranging gives:
\[
-6A - 2A - 8B + 25 - 1 = 0
\]
\[
-8A - 8B + 24 = 0
\]
Dividing by -8:
\[
A + B - 3 = 0 \quad \text{(Equation 1)}
\]
### Step 7: Find the Slope of the Line
The line \(x + y - 7 = 0\) has a slope \(M_2 = -1\). The slope of the radius to the point of tangency \((3, 4)\) from the center \((A, B)\) is:
\[
M_1 = \frac{B - 4}{A - 3}
\]
### Step 8: Use the Perpendicular Condition
Since the radius is perpendicular to the tangent line:
\[
M_1 \cdot M_2 = -1
\]
This gives:
\[
\frac{B - 4}{A - 3} \cdot (-1) = -1
\]
Thus:
\[
B - 4 = A - 3
\]
Rearranging gives:
\[
B = A + 1 \quad \text{(Equation 2)}
\]
### Step 9: Substitute Equation 2 into Equation 1
Substituting \(B = A + 1\) into Equation 1:
\[
A + (A + 1) - 3 = 0
\]
This simplifies to:
\[
2A - 2 = 0
\]
Thus:
\[
A = 1
\]
### Step 10: Find B
Substituting \(A = 1\) into Equation 2:
\[
B = 1 + 1 = 2
\]
### Step 11: Find the Radius
The center of the circle is \((1, 2)\). Now we find the radius by calculating the distance from \((1, 2)\) to \((3, 4)\):
\[
r = \sqrt{(3 - 1)^2 + (4 - 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]
### Step 12: Write the Equation of the Circle
The standard equation of a circle is:
\[
(x - A)^2 + (y - B)^2 = r^2
\]
Substituting \(A = 1\), \(B = 2\), and \(r = 2\sqrt{2}\):
\[
(x - 1)^2 + (y - 2)^2 = (2\sqrt{2})^2
\]
This simplifies to:
\[
(x - 1)^2 + (y - 2)^2 = 8
\]
### Final Equation
Expanding this gives:
\[
x^2 - 2x + 1 + y^2 - 4y + 4 - 8 = 0
\]
Thus:
\[
x^2 + y^2 - 2x - 4y - 3 = 0
\]
### Conclusion
The equation of the circle is:
\[
x^2 + y^2 - 2x - 4y - 3 = 0
\]
