Obtain the equation of the circle, radius 2 units, which lies in the positive quadrant and touches both axes of co-ordinates. Find also the equation of the circle, with centre (6, 5) which touches the above circle externally.

Answer

To solve the problem, we need to find the equations of two circles based on the given conditions.

### Step 1: Find the equation of the first circle
1. **Identify the center and radius**: The radius of the circle is given as 2 units, and since it lies in the positive quadrant and touches both axes, its center will be at (2, 2).

2. **Write the equation of the circle**: The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius. Substituting \(h = 2\), \(k = 2\), and \(r = 2\):
\[ (x - 2)^2 + (y - 2)^2 = 2^2 \]
This simplifies to:
\[ (x - 2)^2 + (y - 2)^2 = 4 \]

3. **Expand the equation**:
\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
Combining these gives:
\[ x^2 - 4x + 4 + y^2 - 4y + 4 = 4 \]
Simplifying further:
\[ x^2 + y^2 - 4x - 4y + 8 = 4 \]
\[ x^2 + y^2 - 4x - 4y + 4 = 0 \]

### Step 2: Find the equation of the second circle
1. **Identify the center of the second circle**: The center is given as (6, 5).

2. **Determine the radius**: The radius of the second circle is the distance from its center (6, 5) to the first circle's center (2, 2) minus the radius of the first circle (2).
\[ \text{Distance} = \sqrt{(6 - 2)^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
Therefore, the radius of the second circle is:
\[ 5 - 2 = 3 \]

3. **Write the equation of the second circle**: Using the center (6, 5) and radius 3:
\[ (x - 6)^2 + (y - 5)^2 = 3^2 \]
This simplifies to:
\[ (x - 6)^2 + (y - 5)^2 = 9 \]

4. **Expand the equation**:
\[ (x - 6)^2 = x^2 - 12x + 36 \]
\[ (y - 5)^2 = y^2 - 10y + 25 \]
Combining these gives:
\[ x^2 - 12x + 36 + y^2 - 10y + 25 = 9 \]
Simplifying further:
\[ x^2 + y^2 - 12x - 10y + 61 = 9 \]
\[ x^2 + y^2 - 12x - 10y + 52 = 0 \]

### Final Equations
1. The equation of the first circle is:
\[ x^2 + y^2 - 4x - 4y + 4 = 0 \]

2. The equation of the second circle is:
\[ x^2 + y^2 - 12x - 10y + 52 = 0 \]