If the line y=2-x is tangent to the circle S at the point P(1,1) and circle S is orthogonal to the circle `x^2+y^2+2x+2y-2=0` then find the length of tangent drawn from the point (2,2) to the circle S

Answer

To solve the problem step-by-step, we will follow the given information and derive the necessary equations to find the length of the tangent drawn from the point (2,2) to the circle S.

### Step 1: Identify the circles and their equations

We have two circles:
1. Circle S (unknown equation)
2. Circle T given by the equation \( x^2 + y^2 + 2x + 2y - 2 = 0 \)

### Step 2: Rewrite Circle T in standard form

To rewrite Circle T in standard form, we complete the square:

\[ x^2 + 2x + y^2 + 2y - 2 = 0 \]

Completing the square for \(x\) and \(y\):

\[ (x+1)^2 - 1 + (y+1)^2 - 1 - 2 = 0 \]

This simplifies to:

\[ (x+1)^2 + (y+1)^2 = 4 \]

So, Circle T has center \((-1, -1)\) and radius \(2\).

### Step 3: Use the orthogonality condition

Circle S is orthogonal to Circle T. The condition for orthogonality of two circles is given by:

\[ 2(g_1g_2 + f_1f_2) = c_1 + c_2 \]

For Circle T, we have:
- \(g_1 = 1\), \(f_1 = 1\), \(c_1 = -2\)

Let Circle S be represented as:

\[ x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 \]

The orthogonality condition becomes:

\[ 2(1 \cdot g_2 + 1 \cdot f_2) = -2 + c_2 \]

### Step 4: Point P lies on Circle S

The point \(P(1,1)\) lies on Circle S, so substituting \(x=1\) and \(y=1\) into the equation of Circle S gives us:

\[ 1^2 + 1^2 + 2g_2(1) + 2f_2(1) + c_2 = 0 \]

This simplifies to:

\[ 2 + 2g_2 + 2f_2 + c_2 = 0 \]

### Step 5: Tangent condition

The line \(y = 2 - x\) is tangent to Circle S at point \(P(1,1)\). The equation of the tangent line can be expressed in point-slope form:

\[ y - 1 = -1(x - 1) \implies x + y - 2 = 0 \]

### Step 6: Set up the equations

From the orthogonality condition:

\[ 2(g_2 + f_2) = -2 + c_2 \quad \text{(1)} \]

From the point \(P(1,1)\):

\[ 2 + 2g_2 + 2f_2 + c_2 = 0 \quad \text{(2)} \]

From the tangent condition, we can equate coefficients:

\[ \frac{1 + g_2}{1} = \frac{1 + f_2}{1} = \frac{g_2 + f_2}{-2} \]

### Step 7: Solve the equations

From (1), we can express \(c_2\):

\[ c_2 = 2(g_2 + f_2) + 2 \]

Substituting \(c_2\) into (2):

\[ 2 + 2g_2 + 2f_2 + (2(g_2 + f_2) + 2) = 0 \]

This simplifies to:

\[ 4 + 4g_2 + 4f_2 = 0 \implies g_2 + f_2 = -1 \]

### Step 8: Find \(g_2\) and \(f_2\)

Now we have two equations:

1. \(g_2 + f_2 = -1\)
2. \(2(g_2 + f_2) = -2 + c_2\)

From the first equation, we can express \(f_2\) in terms of \(g_2\):

\[ f_2 = -1 - g_2 \]

Substituting into the second equation gives:

\[ 2(-1) = -2 + c_2 \implies c_2 = 0 \]

### Step 9: Final equation of Circle S

Now substituting \(c_2 = 0\) back into the equations gives us:

\[ g_2 + f_2 = -1 \quad \text{and} \quad c_2 = 0 \]

Thus, Circle S can be expressed as:

\[ x^2 + y^2 - x - y = 0 \]

### Step 10: Length of tangent from point (2,2)

To find the length of the tangent from point (2,2) to Circle S, we use the formula:

\[ \text{Length} = \sqrt{s_1} \]

Where \(s_1\) is calculated as:

\[ s_1 = x^2 + y^2 + 2g_2x + 2f_2y + c_2 \]

Substituting \(x = 2\), \(y = 2\), \(g_2 = -\frac{1}{2}\), \(f_2 = -\frac{1}{2}\), and \(c_2 = 0\):

\[ s_1 = 2^2 + 2^2 + 2(-\frac{1}{2})(2) + 2(-\frac{1}{2})(2) + 0 \]

Calculating:

\[ s_1 = 4 + 4 - 2 - 2 = 4 \]

Thus, the length of the tangent is:

\[ \text{Length} = \sqrt{4} = 2 \]

### Final Answer

The length of the tangent drawn from the point (2,2) to Circle S is **2**.