A variable circle always touches the line y-x=0 and passes though the point (0,0), the common chords of above cirlce and `x^2+y^2+6x+8y-7=0` will pass through fixed point , whose coordinates are

Options

A. (1,1)
B. (-1,1/2)
C. `(1/2,1/2)`
D. `(1/2,-1)`

Answer

To solve the problem step by step, we need to find the fixed point through which the common chords of the variable circle and the given circle pass.

### Step 1: Understand the given conditions
The variable circle touches the line \(y - x = 0\) (which is the line \(y = x\)) and passes through the point \((0, 0)\).

### Step 2: Write the equation of the variable circle
The general equation of a circle that touches the line \(y = x\) and passes through the origin can be expressed as:
\[ x^2 + y^2 + \lambda(y - x) = 0 \]
where \(\lambda\) is a parameter that determines the specific circle.

### Step 3: Write the equation of the second circle
The second circle is given by:
\[ x^2 + y^2 + 6x + 8y - 7 = 0 \]

### Step 4: Find the equation of the common chord
The common chord of the two circles can be found by subtracting the equations of the circles:
\[ S_1 - S_2 = 0 \]
Substituting the equations:
\[ (x^2 + y^2 + \lambda(y - x)) - (x^2 + y^2 + 6x + 8y - 7) = 0 \]
This simplifies to:
\[ \lambda(y - x) - (6x + 8y - 7) = 0 \]
Rearranging gives:
\[ \lambda y - \lambda x - 6x - 8y + 7 = 0 \]
Combining like terms results in:
\[ (-\lambda - 6)x + (\lambda - 8)y + 7 = 0 \]

### Step 5: Analyze the equation of the common chord
The equation of the common chord can be expressed in the form:
\[ A x + B y + C = 0 \]
where \(A = -\lambda - 6\), \(B = \lambda - 8\), and \(C = 7\).

### Step 6: Determine the fixed point
For the common chords to pass through a fixed point, the coefficients of \(x\) and \(y\) must satisfy a certain condition. This means that the lines represented by the common chords must be dependent on each other.

Setting up the equations:
1. \(y = x\) (from the line \(y - x = 0\))
2. The equation of the common chord.

Substituting \(y = x\) into the common chord equation:
\[ (-\lambda - 6)x + (\lambda - 8)x + 7 = 0 \]
This simplifies to:
\[ (-\lambda - 6 + \lambda - 8)x + 7 = 0 \]
\[ (-14)x + 7 = 0 \]
This implies:
\[ x = \frac{1}{2} \]
Since \(y = x\), we also have:
\[ y = \frac{1}{2} \]

### Final Answer
Thus, the fixed point through which the common chords pass is:
\[ \left(\frac{1}{2}, \frac{1}{2}\right) \]