Find the equation of the circle having
centre at (3,2), and touching the line
` 4x + 3y - 8 = 0` .

Answer

To find the equation of the circle with center at (3, 2) that touches the line \(4x + 3y - 8 = 0\), we will follow these steps:

### Step 1: Identify the center of the circle and the line equation
The center of the circle is given as \(C(3, 2)\) and the equation of the line is \(4x + 3y - 8 = 0\).

### Step 2: Calculate the perpendicular distance from the center to the line
The distance \(d\) from a point \((H, K)\) to a line \(Ax + By + C = 0\) is given by the formula:
\[ d = \frac{|AH + BK + C|}{\sqrt{A^2 + B^2}} \]
For our line \(4x + 3y - 8 = 0\), we have:
- \(A = 4\)
- \(B = 3\)
- \(C = -8\)

Substituting the center of the circle \(C(3, 2)\) into the formula:
\[ d = \frac{|4(3) + 3(2) - 8|}{\sqrt{4^2 + 3^2}} = \frac{|12 + 6 - 8|}{\sqrt{16 + 9}} = \frac{|10|}{\sqrt{25}} = \frac{10}{5} = 2 \]
Thus, the radius \(r\) of the circle is \(2\).

### Step 3: Write the equation of the circle
The equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 3\), \(k = 2\), and \(r = 2\):
\[ (x - 3)^2 + (y - 2)^2 = 2^2 \]
This simplifies to:
\[ (x - 3)^2 + (y - 2)^2 = 4 \]

### Step 4: Expand the equation
Expanding the equation:
\[ (x - 3)^2 = x^2 - 6x + 9 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
Combining these gives:
\[ x^2 - 6x + 9 + y^2 - 4y + 4 = 4 \]
Simplifying further:
\[ x^2 + y^2 - 6x - 4y + 13 - 4 = 0 \]
\[ x^2 + y^2 - 6x - 4y + 9 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 6x - 4y + 9 = 0 \]

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