If `x , y in R` satisfies `(x+5)^2+(y-12)^2=(14)^2,` then the minimum value of `sqrt(x^2=y^2)` is__________

Explanation

To solve the problem, we need to find the minimum value of \( \sqrt{x^2 + y^2} \) given the constraint \( (x + 5)^2 + (y - 12)^2 = 14^2 \).

### Step-by-Step Solution

1. **Understand the Constraint**:
The equation \( (x + 5)^2 + (y - 12)^2 = 14^2 \) represents a circle centered at the point \((-5, 12)\) with a radius of \(14\).

2. **Express \(x\) and \(y\)**:
We can parametrize the points on the circle using trigonometric functions:
\[ x + 5 = 14 \cos \theta \quad \text{and} \quad y - 12 = 14 \sin \theta \]
Therefore, we can express \(x\) and \(y\) as:
\[ x = 14 \cos \theta - 5 \quad \text{and} \quad y = 14 \sin \theta + 12 \]

3. **Substitute into \(x^2 + y^2\)**:
Now, we substitute \(x\) and \(y\) into the expression \(x^2 + y^2\):
\[ x^2 + y^2 = (14 \cos \theta - 5)^2 + (14 \sin \theta + 12)^2 \]

4. **Expand the Squares**:
Expanding both squares gives:
\[ x^2 = (14 \cos \theta - 5)^2 = 196 \cos^2 \theta - 140 \cos \theta + 25 \]
\[ y^2 = (14 \sin \theta + 12)^2 = 196 \sin^2 \theta + 336 \sin \theta + 144 \]

5. **Combine the Expressions**:
Now, combine \(x^2\) and \(y^2\):
\[ x^2 + y^2 = 196 (\cos^2 \theta + \sin^2 \theta) - 140 \cos \theta + 336 \sin \theta + 25 + 144 \]
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ x^2 + y^2 = 196 - 140 \cos \theta + 336 \sin \theta + 169 \]
\[ x^2 + y^2 = 365 - 140 \cos \theta + 336 \sin \theta \]

6. **Find the Minimum Value**:
To minimize \(x^2 + y^2\), we need to minimize the expression \(-140 \cos \theta + 336 \sin \theta\). This can be rewritten in the form \(R \sin(\theta + \phi)\) where:
\[ R = \sqrt{(-140)^2 + (336)^2} = \sqrt{19600 + 112896} = \sqrt{132496} = 364 \]
The angle \(\phi\) can be found using:
\[ \tan \phi = \frac{336}{-140} \]
The minimum value of \(-R\) occurs when \(\sin(\theta + \phi) = -1\), giving us:
\[ \text{Minimum of } (-140 \cos \theta + 336 \sin \theta) = -364 \]

7. **Calculate the Minimum of \(x^2 + y^2\)**:
Thus, the minimum value of \(x^2 + y^2\) is:
\[ 365 - 364 = 1 \]

8. **Final Result**:
Therefore, the minimum value of \(\sqrt{x^2 + y^2}\) is:
\[ \sqrt{1} = 1 \]

### Final Answer:
The minimum value of \(\sqrt{x^2 + y^2}\) is **1**.