Figure 3 shows a circle C with centre Q and radius 4 and the point T which lies on C - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 1

To find the value of k, we can use the Pythagorean theorem. From the information given:

1. The distance from the origin O(0, 0) to the center of the circle Q(11, k) can be expressed as:

   $OQ = \sqrt{(11 - 0)^2 + (k - 0)^2} = \sqrt{11^2 + k^2} = \sqrt{121 + k^2}$

2. We also know that the radius of the circle is 4, and the distance OT is given as $OT = 6\sqrt{5}$.

3. Applying the tangent-secant theorem, we know that:

   $OQ^2 = OT^2 + r^2$

   Substituting the values gives:

   $(\sqrt{121 + k^2})^2 = (6\sqrt{5})^2 + 4^2$

   $121 + k^2 = 180 + 16$

   $121 + k^2 = 196$

   $k^2 = 196 - 121$

   $k^2 = 75$

   Thus, taking the positive root as k is a positive constant:

   $k = \sqrt{75} = 5\sqrt{3}$