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

Answer: (B)

$\frac{5}{13}$

Answer 2

Given four distinct points $(2k, 3k)$, $(1, 0)$, $(0, 1)$, and $(0, 0)$ lie on a circle. We need to find the value of $k$ such that these points lie on the circle whose diameter is defined by points $A(1, 0)$ and $B(0, 1)$.

Step 1. Equation of the Circle: The general equation of a circle with diameter $AB$ is given by:
$(x - 1)(x) + (y - 1)(y) = 0$

Expanding this gives:

$x^2 + y^2 - x - y = 0 \quad \text{...(i)}$

Step 2. Substituting Point $(2k, 3k)$ into the Circle’s Equation: To satisfy the equation, substitute $x = 2k$ and $y = 3k$ into equation (i):

$(2k)^2 + (3k)^2 - 2k - 3k = 0$

Simplifying:

$4k^2 + 9k^2 - 2k - 3k = 0$

$13k^2 - 5k = 0$

Factoring:

$k(13k - 5) = 0$

Therefore, the possible values of $k$ are:

$k = 0 \quad \text{or} \quad k = \frac{5}{13}$

Step 3. Validating the Value of $k$: Since $k = 0$ does not represent a distinct point, we have: $k = \frac{5}{13}$

Answer: $\left( 2 \right) \frac{5}{13}$