4. The equation is given (a') Prove that the above equation represents a circle \( C \) whose center and radius are to be found. (b) Prove that the line \( \varepsilon: y=x-3 \) is tangent to the circle \( C \) and then find their point of contact.

To show that the given equation represents a circle, you can rewrite it in the standard form \((x - h)^2 + (y - k)^2 = r^2\). This typically involves rearranging the equation and completing the square for the \(x\) and \(y\) terms. Once in that form, \( (h, k) \) gives you the center of the circle, and \( r \) is the radius determined by taking the square root of the constant on the right side. To prove that the line \( \varepsilon: y=x-3 \) is tangent to the circle \( C \), you need to find the point of intersection by substituting \( y \) from the line equation into the circle equation. After solving for \(x\), if there's exactly one solution, the line is tangent. For the point of contact, substitute the resulting \(x\) back into \(y = x - 3\) to find the \(y\)-coordinate. Voilà, you've located the point where the line just kisses the circle!