The point at which the circles `x^(2)+y^(2)-4x-4y+7=0 and x^(2)+y^(2)-12x-10y+45=0` touch each other is

Options

A. `(13/5,14/5)`
B. `(2/5,5/6)`
C. `(14/5, 13/5)`
D. `(12/5, 2+(sqrt(21))/4)`

Answer

To find the point at which the circles \(x^2 + y^2 - 4x - 4y + 7 = 0\) and \(x^2 + y^2 - 12x - 10y + 45 = 0\) touch each other, we will follow these steps:

### Step 1: Identify the centers and radii of the circles

1. **Circle 1**: The equation is \(x^2 + y^2 - 4x - 4y + 7 = 0\).
- Rearranging gives:
\[ (x^2 - 4x) + (y^2 - 4y) + 7 = 0 \]
- Completing the square:
\[ (x - 2)^2 - 4 + (y - 2)^2 - 4 + 7 = 0 \implies (x - 2)^2 + (y - 2)^2 = 1 \]
- Center \(C_1 = (2, 2)\) and radius \(r_1 = 1\).

2. **Circle 2**: The equation is \(x^2 + y^2 - 12x - 10y + 45 = 0\).
- Rearranging gives:
\[ (x^2 - 12x) + (y^2 - 10y) + 45 = 0 \]
- Completing the square:
\[ (x - 6)^2 - 36 + (y - 5)^2 - 25 + 45 = 0 \implies (x - 6)^2 + (y - 5)^2 = 16 \]
- Center \(C_2 = (6, 5)\) and radius \(r_2 = 4\).

### Step 2: Find the ratio of the distances from the point of tangency to the centers

Since the circles touch each other externally, the ratio of the distances from the point of tangency \(P\) to the centers \(C_1\) and \(C_2\) is given by the ratio of their radii:
\[ \frac{C_1P}{C_2P} = \frac{r_1}{r_2} = \frac{1}{4} \]

### Step 3: Use the section formula to find the coordinates of point \(P\)

Let \(P\) divide the line segment \(C_1C_2\) in the ratio \(1:4\). The coordinates of \(P\) can be calculated using the section formula:
\[ P\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \]
where \(C_1 = (2, 2)\) and \(C_2 = (6, 5)\).

Substituting \(m = 1\), \(n = 4\), \(x_1 = 2\), \(y_1 = 2\), \(x_2 = 6\), and \(y_2 = 5\):
\[ P_x = \frac{1 \cdot 6 + 4 \cdot 2}{1 + 4} = \frac{6 + 8}{5} = \frac{14}{5} = 2.8 \]
\[ P_y = \frac{1 \cdot 5 + 4 \cdot 2}{1 + 4} = \frac{5 + 8}{5} = \frac{13}{5} = 2.6 \]

Thus, the coordinates of point \(P\) are \(\left(2.8, 2.6\right)\).

### Final Answer:
The point at which the circles touch each other is \(\left(2.8, 2.6\right)\).