Let `S={(x,y)|x,y in R, x^2+y^2-10x+16=0}`. The largest value of `y/x` can be put in the form of `m/n` where m, n are relatively prime natural numbers, then `m^2+n^2=`

Answer

To solve the problem, we need to find the largest value of \( \frac{y}{x} \) for the points on the circle defined by the equation \( x^2 + y^2 - 10x + 16 = 0 \).

### Step 1: Rewrite the equation of the circle
We start by rewriting the equation in standard form. The given equation is:

\[ x^2 + y^2 - 10x + 16 = 0 \]

We can rearrange it as:

\[ x^2 - 10x + y^2 + 16 = 0 \]

### Step 2: Complete the square for \( x \)
To complete the square for the \( x \) terms:

\[ x^2 - 10x = (x - 5)^2 - 25 \]

Substituting this back into the equation gives:

\[ (x - 5)^2 - 25 + y^2 + 16 = 0 \]

This simplifies to:

\[ (x - 5)^2 + y^2 - 9 = 0 \]

### Step 3: Write the equation in standard circle form
Rearranging gives us the standard form of the circle:

\[ (x - 5)^2 + y^2 = 9 \]

This indicates that the center of the circle is at \( (5, 0) \) and the radius is \( 3 \).

### Step 4: Find the slope of the line through the origin
Let \( \frac{y}{x} = k \), then we can express \( y \) in terms of \( x \):

\[ y = kx \]

This line passes through the origin and has a slope \( k \).

### Step 5: Determine the condition for tangency
For the line \( y = kx \) to be tangent to the circle, the distance from the center of the circle \( (5, 0) \) to the line must equal the radius.

The distance \( d \) from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For our line \( y = kx \), we can rewrite it as:

\[ -kx + y = 0 \quad \Rightarrow \quad A = -k, B = 1, C = 0 \]

Thus, the distance from the center \( (5, 0) \) to the line is:

\[ d = \frac{|-k(5) + 1(0) + 0|}{\sqrt{(-k)^2 + 1^2}} = \frac{5k}{\sqrt{k^2 + 1}} \]

### Step 6: Set the distance equal to the radius
Setting this equal to the radius \( 3 \):

\[ \frac{5k}{\sqrt{k^2 + 1}} = 3 \]

### Step 7: Solve for \( k \)
Cross-multiplying gives:

\[ 5k = 3\sqrt{k^2 + 1} \]

Squaring both sides:

\[ 25k^2 = 9(k^2 + 1) \]

Expanding and rearranging:

\[ 25k^2 = 9k^2 + 9 \quad \Rightarrow \quad 16k^2 = 9 \quad \Rightarrow \quad k^2 = \frac{9}{16} \]

Thus,

\[ k = \frac{3}{4} \quad \text{(since we want the largest value)} \]

### Step 8: Write \( k \) in the form \( \frac{m}{n} \)
Here, \( k = \frac{3}{4} \), where \( m = 3 \) and \( n = 4 \).

### Step 9: Find \( m^2 + n^2 \)
Now we need to find \( m^2 + n^2 \):

\[ m^2 + n^2 = 3^2 + 4^2 = 9 + 16 = 25 \]

### Final Answer
Thus, the final answer is:

\[ \boxed{25} \]