States (−4, 6) Follow through their squared brackets from part 9(a)
B1F
1.2
(1)
Typical solution
(−4, 6)
Mark scheme (c)
Scheme
Marks
AO
Substitutes \(x = 0\) correctly into the given or their equation of circle to form a quadratic equation in \(y\) or uses their right-angled triangle correctly to find vertical distance of \(Q\) above \(P\) PI by \(y = 9\) or (0, 9) May be seen on diagram
M1
3.1a
Obtains 9 Accept (0, 9) Must come from centre (−4, 6)
A1
2.2a
(2)
Typical solution
\[4^2 + (y - 6)^2 = 25\]\[y^2 - 12y + 27 = 0\]\[y = 3 \text{ or } y = 9\]\[\therefore y = 9\]
Mark scheme (d)
Scheme
Marks
AO
Uses distance formula to find \(QR\) or \(PR\) PI by their 15 for \(QR\) or their \(5\sqrt{10}\) for \(PR\) OE PI by correct answer or AWFW [71.5, 72]°
M1
1.1a
Substitutes all their relevant lengths correctly into an appropriate trigonometric equation PI by correct answer or AWFW [71.5, 72]°