Find the equation to the circle which
touches each axis at a distance 5 from the origin.

Answer

To find the equation of the circle that touches each axis at a distance of 5 from the origin, we can follow these steps:

### Step 1: Understand the Position of the Circle
Since the circle touches each axis at a distance of 5 from the origin, the points of tangency will be (5, 0), (0, 5), (-5, 0), and (0, -5). This means that the radius of the circle is 5.

### Step 2: Determine the Center of the Circle
The center of the circle must be located at a point that is 5 units away from both the x-axis and the y-axis. Therefore, if we consider the first quadrant, the center of the circle will be at (5, 5). However, the circle can also be in other quadrants.

- In the second quadrant, the center would be (-5, 5).
- In the third quadrant, the center would be (-5, -5).
- In the fourth quadrant, the center would be (5, -5).

### Step 3: Write the Standard Equation of the Circle
The standard equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
For our case, let's take the center at (5, 5) and radius 5. Thus, we have:
\[ (x - 5)^2 + (y - 5)^2 = 5^2 \]
This simplifies to:
\[ (x - 5)^2 + (y - 5)^2 = 25 \]

### Step 4: Expand the Equation
Now, let's expand the equation:
\[ (x - 5)^2 + (y - 5)^2 = 25 \]
Expanding both squares:
\[ (x^2 - 10x + 25) + (y^2 - 10y + 25) = 25 \]
Combining like terms:
\[ x^2 + y^2 - 10x - 10y + 50 = 25 \]
Now, subtract 25 from both sides:
\[ x^2 + y^2 - 10x - 10y + 25 = 0 \]

### Final Equation
Thus, the equation of the circle that touches each axis at a distance of 5 from the origin is:
\[ x^2 + y^2 - 10x - 10y + 25 = 0 \]

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