Question

Four distinct points \( (2k, 3k), (1, 0), (0, 1) \) and \( (0, 0) \) lie on a circle for \( k \) equal to:

• A. \(\frac{2}{13}\)
• B. \(\frac{5}{13}\)
• C. \(\frac{1}{13}\)
• D. \(\frac{2}{13}\)

Answer 1

The Correct Option is B

To find the value of \( k \) for which the points \( (2k, 3k), (1, 0), (0, 1), \) and \( (0, 0) \) lie on the same circle, we apply the general equation of a circle: \(x^2 + y^2 + 2gx + 2fy + c = 0\).

We are given four points: \( A = (2k, 3k), B = (1, 0), C = (0, 1) \), and \( D = (0, 0) \). Concyclicity is determined by a determinant condition. The determinant of the matrix formed by the coordinates and a column of ones must be zero for the points to be concyclic:

x	y	x² + y²	1
2k	3k	\(13k^2\)	1
1	0	1	1
0	1	1	1
0	0	0	1

Evaluating the determinant of this system:

Expanding along the last column:

\(= 1 \cdot \begin{vmatrix} 2k & 3k & 13k^2 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 2k & 3k & 13k^2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2k & 3k & 13k^2 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \end{vmatrix}\)

The first minor expands to:

\((2k)(0 \cdot 1 - 1 \cdot 1) - (3k)(1 \cdot 1 - 0 \cdot 1) + (13k^2)(1 \cdot 1 - 0 \cdot 0)\)

\(= -2k - 3k + 13k^2 = 13k^2 - 5k\)

The other minors evaluate to zero. Setting the determinant equal to zero:

\(13k^2 - 5k = 0\)

Solving for \( k \):

\(k(13k - 5) = 0\)

Thus, \( k = 0 \) or \( k = \frac{5}{13} \). For the points to be distinct and form a circle, \( k eq 0 \). Therefore, \( k = \frac{5}{13} \).