If M and m are the maximum and minimum values of `(y)/(x)` for pair of real numbers (x, y) which satisfy the equation `(x-3)^(2)+(y-3)^(2)=6`, then the value of `(1)/(M)+(1)/(m)` is

Answer

To solve the problem, we need to find the maximum and minimum values of \(\frac{y}{x}\) for the points \((x, y)\) that satisfy the equation of the circle \((x - 3)^2 + (y - 3)^2 = 6\).

### Step 1: Understand the Circle Equation
The given equation \((x - 3)^2 + (y - 3)^2 = 6\) represents a circle with:
- Center at \((3, 3)\)
- Radius \(r = \sqrt{6}\)

### Step 2: Express \(y\) in terms of \(x\)
We can express \(y\) in terms of \(x\) using the equation of the circle:
\[ y - 3 = \sqrt{6 - (x - 3)^2} \quad \text{and} \quad y - 3 = -\sqrt{6 - (x - 3)^2} \]
Thus, we have:
\[ y = 3 + \sqrt{6 - (x - 3)^2} \quad \text{and} \quad y = 3 - \sqrt{6 - (x - 3)^2} \]

### Step 3: Find \(\frac{y}{x}\)
We need to maximize and minimize \(\frac{y}{x}\):
\[ \frac{y}{x} = \frac{3 + \sqrt{6 - (x - 3)^2}}{x} \quad \text{and} \quad \frac{y}{x} = \frac{3 - \sqrt{6 - (x - 3)^2}}{x} \]

### Step 4: Use the Tangent Method
To find the maximum and minimum values of \(\frac{y}{x}\), we can use the concept of tangents from the center of the circle to the lines \(y = kx\) where \(k = \frac{y}{x}\). The maximum and minimum values of \(k\) correspond to the slopes of the tangents to the circle.

### Step 5: Find the Slopes of Tangents
The distance from the center \((3, 3)\) to the line \(y = kx\) must equal the radius \(\sqrt{6}\). The distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(y - kx = 0\) (or \(kx - y + 0 = 0\)), we have \(A = k\), \(B = -1\), and \(C = 0\). The distance from the center \((3, 3)\) is:
\[ d = \frac{|k(3) - 3|}{\sqrt{k^2 + 1}} = \sqrt{6} \]
Setting this equal to the radius:
\[ \frac{|3k - 3|}{\sqrt{k^2 + 1}} = \sqrt{6} \]

### Step 6: Solve for \(k\)
Squaring both sides gives:
\[ (3k - 3)^2 = 6(k^2 + 1) \]
Expanding and simplifying:
\[ 9k^2 - 18k + 9 = 6k^2 + 6 \]
\[ 3k^2 - 18k + 3 = 0 \]
Dividing by 3:
\[ k^2 - 6k + 1 = 0 \]
Using the quadratic formula:
\[ k = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2} \]
Thus, the maximum value \(M = 3 + 2\sqrt{2}\) and the minimum value \(m = 3 - 2\sqrt{2}\).

### Step 7: Calculate \(\frac{1}{M} + \frac{1}{m}\)
Now we calculate:
\[ \frac{1}{M} + \frac{1}{m} = \frac{1}{3 + 2\sqrt{2}} + \frac{1}{3 - 2\sqrt{2}} \]
Finding a common denominator:
\[ = \frac{(3 - 2\sqrt{2}) + (3 + 2\sqrt{2})}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = \frac{6}{9 - 8} = 6 \]

### Final Answer
Thus, the value of \(\frac{1}{M} + \frac{1}{m}\) is:
\[ \boxed{6} \]