If the circle `x^2+y^2+2gx+2fy+c=0` is touched by `y=x` at `P` such that `O P=6sqrt(2),` then the value of `c` is

Options

A. 36
B. 144
C. 72
D. none of these

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Circle and Line Equation
We are given the equation of a circle:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
and the line \( y = x \) touches this circle at point \( P \).

### Step 2: Distance from the Origin to the Line
The distance from the origin \( O(0, 0) \) to the line \( y = x \) can be calculated using the formula for the distance from a point to a line. The line can be rewritten in the standard form as:
\[ x - y = 0 \]
The distance \( d \) from point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line:
- \( A = 1 \)
- \( B = -1 \)
- \( C = 0 \)

Thus, the distance from the origin to the line is:
\[ d = \frac{|1(0) + (-1)(0) + 0|}{\sqrt{1^2 + (-1)^2}} = \frac{0}{\sqrt{2}} = 0 \]

However, since we know that the distance \( OP = 6\sqrt{2} \), we will calculate the coordinates of point \( P \).

### Step 3: Find the Coordinates of Point P
Since \( OP = 6\sqrt{2} \) and the line \( y = x \) implies that \( P \) has coordinates \( (x, x) \), we can express the distance as:
\[ OP = \frac{|x - x|}{\sqrt{1^2 + (-1)^2}} = \frac{|0|}{\sqrt{2}} = 0 \]
This means we need to find the coordinates using the distance formula:
\[ OP = \sqrt{x^2 + x^2} = \sqrt{2x^2} = x\sqrt{2} \]
Setting this equal to \( 6\sqrt{2} \):
\[ x\sqrt{2} = 6\sqrt{2} \]
Dividing both sides by \( \sqrt{2} \):
\[ x = 6 \]
Thus, the coordinates of point \( P \) are \( (6, 6) \).

### Step 4: Substitute into the Circle Equation
Since the line touches the circle, the point \( P(6, 6) \) must satisfy the circle's equation:
\[ 6^2 + 6^2 + 2g(6) + 2f(6) + c = 0 \]
Simplifying this gives:
\[ 36 + 36 + 12g + 12f + c = 0 \]
\[ 72 + 12g + 12f + c = 0 \]
This can be rearranged as:
\[ 12g + 12f + c = -72 \]
(Equation 1)

### Step 5: Condition for Tangency
For the line \( y = x \) to be tangent to the circle, the quadratic equation formed by substituting \( y = x \) into the circle's equation must have equal roots. We substitute \( y = x \) into the circle's equation:
\[ x^2 + x^2 + 2gx + 2fx + c = 0 \]
This simplifies to:
\[ 2x^2 + (2g + 2f)x + c = 0 \]
For this quadratic to have equal roots, the discriminant must be zero:
\[ (2g + 2f)^2 - 4(2)(c) = 0 \]
This simplifies to:
\[ (g + f)^2 = 2c \]
(Equation 2)

### Step 6: Solve the Equations
From Equation 1:
\[ c = -72 - 12g - 12f \]
Substituting into Equation 2:
\[ (g + f)^2 = 2(-72 - 12g - 12f) \]
Expanding this gives:
\[ (g + f)^2 = -144 - 24g - 24f \]

### Step 7: Rearranging and Solving
Let \( s = g + f \):
\[ s^2 + 24s + 144 = 0 \]
This can be factored as:
\[ (s + 12)^2 = 0 \]
Thus, \( s = -12 \), which means:
\[ g + f = -12 \]

### Step 8: Substitute Back to Find c
Substituting \( g + f = -12 \) back into Equation 2:
\[ c = -72 - 12(-12) = -72 + 144 = 72 \]

### Final Answer
Thus, the value of \( c \) is:
\[ \boxed{72} \]