Prove that the staraight line ` y = x + c sqrt"" 2 ` touches the circle ` x^(2) + y^(2) = c^(2)` and find its point of contact

Answer

To prove that the straight line \( y = x + c\sqrt{2} \) touches the circle \( x^2 + y^2 = c^2 \) and to find the point of contact, we will follow these steps:

### Step 1: Substitute the equation of the line into the equation of the circle.

We start with the circle's equation:
\[ x^2 + y^2 = c^2 \]
Substituting \( y = x + c\sqrt{2} \) into the circle's equation:
\[ x^2 + (x + c\sqrt{2})^2 = c^2 \]

### Step 2: Expand the equation.

Expanding \( (x + c\sqrt{2})^2 \):
\[ x^2 + (x^2 + 2xc\sqrt{2} + 2c^2) = c^2 \]
This simplifies to:
\[ x^2 + x^2 + 2xc\sqrt{2} + 2c^2 = c^2 \]
Combining like terms:
\[ 2x^2 + 2xc\sqrt{2} + 2c^2 - c^2 = 0 \]
This simplifies to:
\[ 2x^2 + 2xc\sqrt{2} + c^2 = 0 \]
Dividing the entire equation by 2:
\[ x^2 + xc\sqrt{2} + \frac{c^2}{2} = 0 \]

### Step 3: Determine the discriminant.

For the line to touch the circle, the quadratic equation must have exactly one solution. This occurs when the discriminant is zero. The discriminant \( D \) of the quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
In our case:
- \( a = 1 \)
- \( b = c\sqrt{2} \)
- \( c = \frac{c^2}{2} \)

Calculating the discriminant:
\[ D = (c\sqrt{2})^2 - 4 \cdot 1 \cdot \frac{c^2}{2} \]
\[ D = 2c^2 - 2c^2 = 0 \]
Since the discriminant is zero, the line touches the circle.

### Step 4: Find the point of contact.

To find the point of contact, we can use the quadratic equation we derived:
\[ x^2 + xc\sqrt{2} + \frac{c^2}{2} = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \):
Since \( D = 0 \):
\[ x = \frac{-c\sqrt{2}}{2} \]
Now substituting \( x \) back into the equation of the line to find \( y \):
\[ y = x + c\sqrt{2} = \frac{-c\sqrt{2}}{2} + c\sqrt{2} = \frac{c\sqrt{2}}{2} \]

### Final Result:

Thus, the point of contact is:
\[ \left( -\frac{c\sqrt{2}}{2}, \frac{c\sqrt{2}}{2} \right) \]