If the points (0,0),(2,0),(0,-2) and (k,-2) are concylic then k=

Options

A. 2
B. `-2`
C. 0
D. 1

Answer

To find the value of \( k \) such that the points \( (0,0) \), \( (2,0) \), \( (0,-2) \), and \( (k,-2) \) are concyclic, we can follow these steps:

### Step 1: Understand the condition for concyclic points
Four points are concyclic if they lie on the same circle. This means that the points will satisfy the general equation of a circle, which can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

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

### Step 3: Substitute the second point \( (2,0) \)
Substituting \( (2,0) \) into the equation:
\[ 2^2 + 0^2 + 2g(2) + 2f(0) + c = 0 \]
This simplifies to:
\[ 4 + 4g + 0 + 0 = 0 \implies 4g = -4 \implies g = -1 \]

### Step 4: Substitute the third point \( (0,-2) \)
Substituting \( (0,-2) \) into the equation:
\[ 0^2 + (-2)^2 + 2g(0) + 2f(-2) + c = 0 \]
This simplifies to:
\[ 4 + 0 - 4f + 0 = 0 \implies 4 = 4f \implies f = 1 \]

### Step 5: Write the equation of the circle
Now we have \( g = -1 \), \( f = 1 \), and \( c = 0 \). The equation of the circle becomes:
\[ x^2 + y^2 - 2x + 2y = 0 \]

### Step 6: Substitute the fourth point \( (k,-2) \)
Substituting \( (k,-2) \) into the equation:
\[ k^2 + (-2)^2 - 2k + 2(-2) = 0 \]
This simplifies to:
\[ k^2 + 4 - 2k - 4 = 0 \implies k^2 - 2k = 0 \]

### Step 7: Factor the equation
Factoring gives:
\[ k(k - 2) = 0 \]
This implies:
\[ k = 0 \quad \text{or} \quad k = 2 \]

### Step 8: Conclusion
Since we are looking for the value of \( k \) that makes the points concyclic, we find:
\[ k = 2 \]

Thus, the value of \( k \) is \( 2 \).

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