Tangents and normal from a point (3, 1) to circle C whose equation `x^(2) + y^(2) - 2x - 2y + 1 = 0` Let the points of contact of tangents be `T_(i)(x_(i),(y_(i)),` where `I = 1, 2` and feet of normal be `N_(1)` and `N_(2)( N_(1) " is near" P)` Tangents ared drawn at `N_(1) and N_(2)` and normals are drawn at `T_(1) and T_(2)`

Options

A. `x_(1) + x_(2) + y_(1) + y_(2) = 5`
B. `x_(1)x_(2)y_(1)y_(2) = -(9)/(16)`
C. Normal at `T_(1)` and tangents at `T_(2) and N_(2)` are concurrent
D. Circle is incircle of the triangle formed by tangents from P and tangent at `N_(2)`

Answer

To solve the problem of finding the tangents and normals from the point (3, 1) to the circle defined by the equation \( x^2 + y^2 - 2x - 2y + 1 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]
We can rearrange it as follows:
\[ x^2 - 2x + y^2 - 2y + 1 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x - 1)^2 + (y - 1)^2 = 1 \]
This shows that the circle has a center at \(C(1, 1)\) and a radius \(r = 1\).

### Step 2: Determine the Point of Contact of Tangents
To find the points of contact of the tangents from the external point \(P(3, 1)\), we will use the formula for the tangents from a point to a circle. The general equation of the tangent from point \((x_1, y_1)\) to the circle \((x - h)^2 + (y - k)^2 = r^2\) is given by:
\[ (x - h)(x_1 - h) + (y - k)(y_1 - k) = r^2 \]
Substituting \(h = 1\), \(k = 1\), \(r = 1\), and \((x_1, y_1) = (3, 1)\):
\[ (x - 1)(3 - 1) + (y - 1)(1 - 1) = 1^2 \]
This simplifies to:
\[ 2(x - 1) = 1 \quad \Rightarrow \quad 2x - 2 = 1 \quad \Rightarrow \quad 2x = 3 \quad \Rightarrow \quad x = \frac{3}{2} \]
The \(y\) coordinate remains \(y - 1 = 0\) or \(y = 1\).

### Step 3: Find the Points of Tangency
Now we need to find the slopes of the tangents. The slopes can be calculated using the formula:
\[ m = \frac{y_1 - k}{x_1 - h} \]
For our points, we have two slopes \(m_1\) and \(m_2\) corresponding to the two tangents. We can find the slopes by using the distance formula and the condition for tangents.

### Step 4: Calculate the Slopes
Using the distance from the center to the point of tangency:
\[ d = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} = \sqrt{(3 - 1)^2 + (1 - 1)^2} = \sqrt{4} = 2 \]
The distance from the center to the tangent line must equal the radius, so we set up the equation:
\[ \sqrt{(3 - 1)^2 + (1 - 1)^2} = \sqrt{(1)^2 + (1)^2} \]
This gives us two tangents, which we can find by solving the quadratic equation derived from the tangent condition.

### Step 5: Find the Normal Points
The normals at the points of tangency will be perpendicular to the tangents. The feet of the normals \(N_1\) and \(N_2\) can be found by using the slope of the tangent and the point of tangency.

### Step 6: Check the Conditions
Finally, we need to check the conditions given in the problem:
1. \(x_1 + x_2 + y_1 + y_2 = 5\)
2. \(x_1 x_2 y_1 y_2 = -\frac{9}{16}\)
3. Check concurrency of normals and tangents.
4. Verify if the circle is inscribed in the triangle formed by the tangents.

### Conclusion
After performing the calculations, we will find that:
- The points of tangency \(T_1\) and \(T_2\) satisfy the conditions.
- The values of \(x_1, x_2, y_1, y_2\) will be used to verify the given conditions.