The coordinates of the points P and Q are (1, 1) and (7, 11) respectively. The line segment PQ forms a diameter of a circle.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [B1] Centre (4, 6) found (midpoint of P and Q).
2. [B1] Radius² = (4-1)² + (6-1)² = 34 (or diameter² = (7-1)² + (11-1)² = 136).
3. [M1] Uses their centre (h, k) and radius² r² to form (x-h)² + (y-k)² = r².
4. [A1] (x-4)² + (y-6)² = 34 or x² + y² - 8x - 12y + 18 = 0.
5. [B1] Gradient of PQ = (11-1)/(7-1) = 10/6 = 5/3.
6. [M1] Uses perpendicular gradients: m_tangent = -1/m_PQ.
7. [A1] Equation of tangent: y - 11 = (-3/5)(x - 7) or y = (-3/5)x + 76/5.
8. [B1] (Alternative) Differentiates equation of circle implicitly: 2(x-4) + 2(y-6)dy/dx = 0.
9. [M1] (Alternative) Substitutes (7, 11) into their differentiated equation to find dy/dx.
10. [A1] (Alternative) Equation of tangent: y - 11 = (-3/5)(x - 7) or y = (-3/5)x + 76/5.
11. [B1] Finds y-coordinate of R. (Substitute x=7 into circle equation or by symmetry, R(7,1)).
12. [B1] Finds equation of tangent at R. (Gradient of AR = (6-1)/(4-7) = 5/-3. Tangent at R is perpendicular: m=3/5. y-1 = 3/5(x-7)).
13. [M1] Solves simultaneous equations for the two tangent lines (tangent at Q and tangent at R).
14. [A1] Intersection point (46/3, 6) or equivalent exact form.
15. [B1] (Alternative) y-coordinate of R is 1.
16. [B1] (Alternative) The intersection point is on the horizontal line through the centre of the circle, so y=6.
17. [M1] (Alternative) Substitutes y=6 into the equation of the tangent at Q (or R) and solves for x.
18. [A1] (Alternative) Intersection point (46/3, 6) or equivalent exact form.