Problem Analysis: The question asks for the possible values of k such that the points A(0, 2), B(2, 3), C(4, 5), and D(0, k) all lie on the same circle. We need to find the equation of the circle passing through the first three points and then use the fourth point to solve for k.

Finding the Circle Equation

The general equation of a circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\). We substitute the coordinates of points A, B, and C to form three equations:

1. Point A(0, 2): \(0^2 + 2^2 + 2g(0) + 2f(2) + c = 0 \implies 4 + 4f + c = 0\)
2. Point B(2, 3): \(2^2 + 3^2 + 2g(2) + 2f(3) + c = 0 \implies 13 + 4g + 6f + c = 0\)
3. Point C(4, 5): \(4^2 + 5^2 + 2g(4) + 2f(5) + c = 0 \implies 41 + 8g + 10f + c = 0\)

Solving for Coefficients g, f, and c

Subtracting the equations to eliminate variables:

• (Eq 2) - (Eq 1): \((13 + 4g + 6f + c) - (4 + 4f + c) = 0 \implies 9 + 4g + 2f = 0\) (Eq 4)
• (Eq 3) - (Eq 2): \((41 + 8g + 10f + c) - (13 + 4g + 6f + c) = 0 \implies 28 + 4g + 4f = 0 \implies 7 + g + f = 0\) (Eq 5)

From Eq 5, \(g = -7 - f\). Substitute this into Eq 4:

\(9 + 4(-7 - f) + 2f = 0\)

\(9 - 28 - 4f + 2f = 0\)

\(-19 - 2f = 0 \implies 2f = -19 \implies f = -19/2\)

Now find g using Eq 5:

\(g = -7 - (-19/2) = -7 + 19/2 = (-14 + 19)/2 = 5/2\)

Find c using Eq 1:

\(4 + 4(-19/2) + c = 0 \implies 4 - 38 + c = 0 \implies -34 + c = 0 \implies c = 34\)

The equation of the circle is \(x^2 + y^2 + 5x - 19y + 34 = 0\).

Finding the Value(s) of k

Since point D(0, k) lies on the circle, substitute its coordinates into the circle equation:

\(0^2 + k^2 + 5(0) - 19(k) + 34 = 0\)

\(k^2 - 19k + 34 = 0\)

Solve this quadratic equation for k using the quadratic formula \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(k = \frac{-(-19) \pm \sqrt{(-19)^2 - 4(1)(34)}}{2(1)}\)

\(k = \frac{19 \pm \sqrt{361 - 136}}{2}\)

\(k = \frac{19 \pm \sqrt{225}}{2}\)

\(k = \frac{19 \pm 15}{2}\)

The two possible values for k are:

\(k_1 = \frac{19 + 15}{2} = \frac{34}{2} = 17\)

\(k_2 = \frac{19 - 15}{2} = \frac{4}{2} = 2\)

Therefore, the possible values of k are 2 and 17.