The circle C has equation $x^2 + y^2 - 10x + 4y + 11 = 0$ (a) Find (i) the coordinates of the centre of C, (ii) the exact radius of C, giving your answer as a simplified surd - Edexcel - A-Level Maths Pure - Question 9 - 2021 - Paper 1

Using the standard form of the circle equation $(x - h)^2 + (y - k)^2 = r^2$, where (h, k) is the center and r is the radius, we know: $h = 5, k = -2, r^2 = 18$

To find the radius r, take the square root: $r = ext{sqrt}(18) = 3 ext{sqrt}(2)$

So the exact radius of C is $3 ext{sqrt}(2)$.

Since the line l is a tangent to circle C, the distance from the center of the circle to the line must equal the radius. The equation of the line is given by: $y = 3x + k$

We can rewrite this in the form: $-3x + y - k = 0$

Using the formula for the distance from a point (x₀, y₀) to a line Ax + By + C = 0, the distance D is given by: D = rac{|Ax_0 + By_0 + C|}{ ext{sqrt}(A^2 + B^2)}

In our case:

• A = -3, B = 1, C = -k, (x₀, y₀) is the center of H (5, -2).

Plugging these values in, we can find D: D = rac{|-3(5) + 1(-2) - k|}{ ext{sqrt}((-3)^2 + 1^2)} = rac{| -15 - 2 - k |}{ ext{sqrt}(10)} = rac{| -17 - k |}{ ext{sqrt}(10)}

Setting this distance equal to the radius we found in part (ii), we have: rac{| -17 - k |}{ ext{sqrt}(10)} = 3 ext{sqrt}(2)

Squaring both sides leads to: $(-17 - k)^2 = 18 ext{ (10)}$

Expanding and solving: $k^2 + 34k + 289 = 180$ $k^2 + 34k + 109 = 0$

Using the quadratic formula: k = rac{-b ext{ (±) } ext{sqrt}(b^2 - 4ac)}{2a} where a = 1, b = 34, c = 109. This gives us two potential values for k. Upon simplification, you will find the solutions for k as: $k = -17 ext{ ± } 6 ext{sqrt}(6)$

Thus, the possible values of k are $-17 + 6 ext{sqrt}(6)$ and $-17 - 6 ext{sqrt}(6)$.