Square of diameter of the circle having tangent at `(1,1)` as `x + y -2 =0` and passing through `(2,2)` is ` "_________"`

Answer

To find the square of the diameter of the circle that has a tangent at the point (1,1) given by the equation \(x + y - 2 = 0\) and passes through the point (2,2), we can follow these steps:

### Step 1: Identify the center of the circle
Since the tangent line touches the circle at the point (1,1), the radius at that point is perpendicular to the tangent line. The slope of the line \(x + y - 2 = 0\) can be found by rewriting it in slope-intercept form:

\[ y = -x + 2 \]

The slope of this line is -1. Therefore, the slope of the radius (which is perpendicular to the tangent) will be the negative reciprocal, which is 1. Thus, the equation of the radius can be written as:

\[ y - 1 = 1(x - 1) \implies y = x \]

### Step 2: Find the intersection point of the radius and the line through (2,2)
Now, we need to find the intersection of the line \(y = x\) with the line \(x + y - 2 = 0\). We can substitute \(y = x\) into the tangent line equation:

\[ x + x - 2 = 0 \implies 2x - 2 = 0 \implies x = 1 \]

Substituting back to find \(y\):

\[ y = 1 \]

Thus, the intersection point is (1,1), which we already know is the point of tangency.

### Step 3: Calculate the distance between the center and the point (2,2)
Now, we need to find the distance between the center of the circle (which is also the point (1,1)) and the point (2,2). Using the distance formula:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
where \((x_1, y_1) = (1, 1)\) and \((x_2, y_2) = (2, 2)\):

\[ d = \sqrt{(2 - 1)^2 + (2 - 1)^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Calculate the diameter of the circle
The diameter \(D\) of the circle is twice the radius. Since the radius is the distance \(d\), we have:

\[ D = 2d = 2\sqrt{2} \]

### Step 5: Find the square of the diameter
Finally, we need to find the square of the diameter:

\[ D^2 = (2\sqrt{2})^2 = 4 \cdot 2 = 8 \]

Thus, the square of the diameter of the circle is **8**.