Vertex of a Shifted Parabola

1. Find the equations of the tangents

The given equation of the parabola is:

$$y^2 = 16(x - 4)$$

Any tangent to a shifted parabola of the form $$y^2 = 4a(x - x_0)$$ with slope $$m$$ is given by:

$$y = m(x - x_0) + \frac{a}{m}$$

For the given parabola, $$a = 4$$ and $$x_0 = 4$$. Thus, the equation of the tangent line is:

$$y = m(x - 4) + \frac{4}{m}$$

Since these tangents are drawn from the origin $$O(0,0)$$, substitute $$x = 0$$ and $$y = 0$$:

$$0 = m(0 - 4) + \frac{4}{m}$$

$$-4m + \frac{4}{m} = 0 \implies m^2 = 1 \implies m = \pm 1$$

Therefore, the slopes of the two tangents are $$m_1 = 1$$ and $$m_2 = -1$$.

2. Determine the points of contact P and Q

The coordinates for the point of contact on a shifted parabola are given by $$\left(x_0 + \frac{a}{m^2}, \frac{2a}{m}\right)$$.

For $$m_1$$ = 1:

$$P = \left(4 + \frac{4}{1^2}, \frac{2(4)}{1}\right) = (8, 8)$$

For $$m_2$$ = -1:

$$Q = \left(4 + \frac{4}{(-1)^2}, \frac{2(4)}{-1}\right) = (8, -8)$$

3. Find the equation of the circumcircle

Let the general equation of the circumcircle passing through the triangle OPQ be:

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

Since it passes through the origin $$O(0,0)$$, we have $$c = 0$$.

Since the triangle is perfectly symmetric across the x-axis (points $$(8,8)$$ and $$(8,-8)$$), the y-coordinate of the center is $$0$$, meaning $$f = 0$$.

Now substitute the point $$P(8,8)$$ into the simplified circle equation $$x^2 + y^2 + 2gx = 0$$:

$$8^2 + 8^2 + 2g(8) = 0$$

$$64 + 64 + 16g = 0$$

$$16g = -128 \implies g = -8$$

Thus, the equation of the circumcircle of the triangle OPQ is:

$$x^2 + y^2 - 16x = 0$$

4. Solve for k

The circumcircle passes through the point $$(k, 4)$$. Substitute these coordinates into the circle's equation:

$$k^2 + 4^2 - 16k = 0$$

$$k^2 - 16k + 16 = 0$$

Using the quadratic formula to solve for $$k$$:

$$k = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(16)}}{2(1)}$$

$$k = \frac{16 \pm \sqrt{256 - 64}}{2} = \frac{16 \pm \sqrt{192}}{2}$$

$$k = \frac{16 \pm 8\sqrt{3}}{2} = 8 \pm 4\sqrt{3}$$

Final Answer

The value of k is $$8 \pm 4\sqrt{3}$$