If the points `(0,0), (1,0), (0,1)` and `(t, t)` are concyclic, then `t` is equal to

Options

A. -1
B. 1
C. 2
D. -2

Explanation

To determine the value of \( t \) such that the points \( (0,0), (1,0), (0,1), (t,t) \) are concyclic, we can follow these steps:

### Step 1: Understand the Concyclic Condition
The points are concyclic if they lie on the same circle. The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \( g, f, c \) are constants.

### Step 2: Substitute the Point \( (0,0) \)
Substituting \( (0,0) \) into the circle equation:
\[ 0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \]
This simplifies to:
\[ c = 0 \]

### Step 3: Substitute the Point \( (1,0) \)
Now substitute \( (1,0) \):
\[ 1^2 + 0^2 + 2g(1) + 2f(0) + c = 0 \]
Substituting \( c = 0 \):
\[ 1 + 2g = 0 \]
From this, we find:
\[ 2g = -1 \quad \Rightarrow \quad g = -\frac{1}{2} \]

### Step 4: Substitute the Point \( (0,1) \)
Next, substitute \( (0,1) \):
\[ 0^2 + 1^2 + 2g(0) + 2f(1) + c = 0 \]
Again substituting \( c = 0 \):
\[ 1 + 2f = 0 \]
From this, we find:
\[ 2f = -1 \quad \Rightarrow \quad f = -\frac{1}{2} \]

### Step 5: Write the Circle Equation
Now we can write the equation of the circle using the values of \( g \) and \( f \):
\[ x^2 + y^2 - x - y = 0 \]

### Step 6: Substitute the Point \( (t,t) \)
Now we substitute \( (t,t) \) into the circle equation:
\[ t^2 + t^2 - t - t = 0 \]
This simplifies to:
\[ 2t^2 - 2t = 0 \]
Factoring out \( 2t \):
\[ 2t(t - 1) = 0 \]

### Step 7: Solve for \( t \)
Setting each factor to zero gives:
\[ 2t = 0 \quad \Rightarrow \quad t = 0 \]
or
\[ t - 1 = 0 \quad \Rightarrow \quad t = 1 \]

### Conclusion
Thus, the values of \( t \) are \( 0 \) and \( 1 \). Since the options provided are \( -1, 1, 2, -2 \), the correct answer is:
\[ \boxed{1} \]