The equation of a circle with centre (4,1) and having 3x+4y-1=0 as tangent is

Options

A. `x^(2)+y^(2)-8x=0`
B. `x^(2)+y^(2)-8x-2y+8=0`
C. `x^(2)+y^(2)-8x+2y+8=0`
D. `x^(2)+y^(2)-8x+4=0`

Answer

To find the equation of a circle with center (4, 1) and tangent line given by the equation \(3x + 4y - 1 = 0\), we will follow these steps:

### Step 1: Identify the center of the circle
The center of the circle is given as \( (h, k) = (4, 1) \).

### Step 2: Write the general equation of the circle
The general equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting the center into the equation, we have:
\[ (x - 4)^2 + (y - 1)^2 = r^2 \]

### Step 3: Find the distance from the center to the tangent line
The distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is given by the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( 3x + 4y - 1 = 0 \), we have:
- \( A = 3 \)
- \( B = 4 \)
- \( C = -1 \)

Now, substituting the center \( (4, 1) \) into the formula:
\[ d = \frac{|3(4) + 4(1) - 1|}{\sqrt{3^2 + 4^2}} \]
Calculating the numerator:
\[ = \frac{|12 + 4 - 1|}{\sqrt{9 + 16}} = \frac{|15|}{\sqrt{25}} = \frac{15}{5} = 3 \]
Thus, the radius \( r = 3 \).

### Step 4: Substitute the radius back into the circle equation
Now that we have the radius, we can substitute \( r = 3 \) back into the circle equation:
\[ (x - 4)^2 + (y - 1)^2 = 3^2 \]
This simplifies to:
\[ (x - 4)^2 + (y - 1)^2 = 9 \]

### Step 5: Expand the equation
Now we will expand the equation:
\[ (x - 4)^2 + (y - 1)^2 = 9 \]
Expanding gives:
\[ (x^2 - 8x + 16) + (y^2 - 2y + 1) = 9 \]
Combining like terms:
\[ x^2 + y^2 - 8x - 2y + 17 = 9 \]
Rearranging gives:
\[ x^2 + y^2 - 8x - 2y + 8 = 0 \]

### Final Equation
Thus, the equation of the circle is:
\[ x^2 + y^2 - 8x - 2y + 8 = 0 \]

---