Equation of a circle S(x,y)=0 , (S(2,3)=16) which touches the line 3x+4y-7=0 at (1,1) is given by

Options

A. `x^(2)+y^(2)+x+2y-5=0`
B. `x^(2)+y^(2)+2x+2y-7=0`
C. `x^(2)+y^(2)+4x-6y+13=0`
D. `x^(2)+y^(2)-4x+6y-7=0`

Answer

To find the equation of the circle \( S(x,y) = 0 \) that touches the line \( 3x + 4y - 7 = 0 \) at the point \( (1,1) \) and satisfies \( S(2,3) = 16 \), we can follow these steps:

### Step 1: Write the general equation of the circle
The general equation of a circle that touches a line at a point can be expressed as:
\[ (x - x_1)^2 + (y - y_1)^2 + \lambda (Ax + By + C) = 0 \]
where \( (x_1, y_1) \) is the point of tangency, and \( Ax + By + C = 0 \) is the equation of the tangent line.

### Step 2: Substitute the point of tangency and the line equation
Given the point of tangency \( (1, 1) \) and the line \( 3x + 4y - 7 = 0 \):
\[ (x - 1)^2 + (y - 1)^2 + \lambda (3x + 4y - 7) = 0 \]

### Step 3: Use the condition \( S(2,3) = 16 \)
Substituting \( (x, y) = (2, 3) \) into the equation:
\[ (2 - 1)^2 + (3 - 1)^2 + \lambda (3(2) + 4(3) - 7) = 16 \]
Calculating each term:
\[ 1^2 + 2^2 + \lambda (6 + 12 - 7) = 16 \]
\[ 1 + 4 + \lambda (11) = 16 \]
\[ 5 + 11\lambda = 16 \]

### Step 4: Solve for \( \lambda \)
Rearranging the equation:
\[ 11\lambda = 16 - 5 \]
\[ 11\lambda = 11 \]
\[ \lambda = 1 \]

### Step 5: Substitute \( \lambda \) back into the circle equation
Now substituting \( \lambda = 1 \) back into the equation:
\[ (x - 1)^2 + (y - 1)^2 + (3x + 4y - 7) = 0 \]

### Step 6: Expand and simplify the equation
Expanding the equation:
\[ (x^2 - 2x + 1) + (y^2 - 2y + 1) + (3x + 4y - 7) = 0 \]
Combining like terms:
\[ x^2 + y^2 - 2x - 2y + 3x + 4y + 1 + 1 - 7 = 0 \]
\[ x^2 + y^2 + x + 2y - 5 = 0 \]

### Final Equation
Thus, the equation of the circle is:
\[ x^2 + y^2 + x + 2y - 5 = 0 \]