If `x+y=2` is a tangent to `x^(2)+y^(2)=2`, then the equation of the tangent at the same point of contact to the circle `x^(2)+y^(2) +3x +3y-8=0` is

Options

A. `x-y=6`
B. `x+y=6`
C. `x+y=2`
D. none

Answer

To solve the problem, we need to find the equation of the tangent to the circle \( x^2 + y^2 + 3x + 3y - 8 = 0 \) at the same point of contact where the line \( x + y = 2 \) is tangent to the circle \( x^2 + y^2 = 2 \).

### Step-by-Step Solution:

1. **Identify the point of tangency**:
The line \( x + y = 2 \) is tangent to the circle \( x^2 + y^2 = 2 \). To find the point of tangency, we substitute \( y = 2 - x \) into the circle's equation:
\[ x^2 + (2 - x)^2 = 2 \]
Expanding this:
\[ x^2 + (4 - 4x + x^2) = 2 \]
\[ 2x^2 - 4x + 4 - 2 = 0 \]
\[ 2x^2 - 4x + 2 = 0 \]
Dividing by 2:
\[ x^2 - 2x + 1 = 0 \]
Factoring:
\[ (x - 1)^2 = 0 \implies x = 1 \]
Substituting \( x = 1 \) back into \( y = 2 - x \):
\[ y = 2 - 1 = 1 \]
Thus, the point of tangency is \( (1, 1) \).

2. **Write the equation of the second circle**:
The second circle is given by:
\[ x^2 + y^2 + 3x + 3y - 8 = 0 \]
We can rewrite this in standard form by completing the square:
\[ (x^2 + 3x) + (y^2 + 3y) = 8 \]
Completing the square for \( x \) and \( y \):
\[ (x + \frac{3}{2})^2 - \frac{9}{4} + (y + \frac{3}{2})^2 - \frac{9}{4} = 8 \]
\[ (x + \frac{3}{2})^2 + (y + \frac{3}{2})^2 = 8 + \frac{9}{2} = \frac{25}{2} \]
So, the center of the circle is \( (-\frac{3}{2}, -\frac{3}{2}) \) and the radius is \( \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} \).

3. **Find the equation of the tangent at the point (1, 1)**:
The general formula for the tangent to a circle \( (x - h)^2 + (y - k)^2 = r^2 \) at point \( (x_1, y_1) \) is:
\[ (x - x_1)(x_1 + h) + (y - y_1)(y_1 + k) = r^2 \]
For our circle, substituting \( (h, k) = (-\frac{3}{2}, -\frac{3}{2}) \), \( (x_1, y_1) = (1, 1) \):
\[ (x - 1)(1 - \frac{3}{2}) + (y - 1)(1 - \frac{3}{2}) = \frac{25}{2} \]
Simplifying:
\[ (x - 1)(-\frac{1}{2}) + (y - 1)(-\frac{1}{2}) = \frac{25}{2} \]
\[ -\frac{1}{2}(x - 1 + y - 1) = \frac{25}{2} \]
Multiplying through by -2:
\[ x + y - 2 = -25 \]
Thus:
\[ x + y = 2 - 25 = -23 \]

4. **Final Equation**:
The equation of the tangent at the point of contact to the circle \( x^2 + y^2 + 3x + 3y - 8 = 0 \) is:
\[ x + y = -23 \]