June 2025 Paper 2 Q9

9 A circle with centre \(C\) has equation

\[(x - 12)^2 + (y - 2)^2 = 100\]

The graph with equation

\[y = |3x - 36| - 8\]

intersects the circle at the points \(A\), \(B\) and \(D\) as shown in the diagram.

Point \(D\) is vertically below point \(C\)

(a) State the coordinates of \(C\) [1 mark]

(b) State the coordinates of \(D\) [1 mark]

Find the value of \(a\)

Fully justify your answer.

[3 marks]

(d)

(i) Find, in radians, the angle \(ADB\)

Give your answer to three significant figures.

[2 marks]

(ii) Hence or otherwise find the length of the minor arc \(AB\)

Give your answer to three significant figures.

[2 marks]

Scheme	Marks	AO
States (12, 2)	B1	1.1b
	(1)

Scheme	Marks	AO
States \((12, -8)\)	R1	2.2a
	(1)

Mark scheme (c)

Scheme	Marks	AO
Forms an equation for \(x\) or \(a\) by Substituting \(y = 10\) into \((x - 12)^2 + (y - 2)^2 = 100\) or \(y = \|3x - 36\| - 8\) Or Substitutes \(\pm(3x - 36) - 8\) for \(y\) in the circle equation Or Forms a right-angled triangle with a hypotenuse of 10 and a shorter side of 8 and uses an appropriate process to find the length of the third side PI by a 6, 8, 10 right-angled triangle	M1	3.1a
Solves their equation to find a value of \(x\) or \(a\) Or Identifies 6, 8 and 10 as a Pythagorean triple PI by a 6, 8, 10 right-angled triangle	M1	1.1a
Completes reasoned argument to obtain \(a = 6\) with a fully correct solution If a 6, 8, 10 approach is used we must see \(12 - 6 = a\), \(12 - 6 = 6\) or \(12 - a = 6\) Accept (6, 10) NMS scores M0M0R0	R1	2.1
	(3)

\[(a - 12)^2 + (10 - 2)^2 = 100\]\[a = 6 \text{ or } 18\]\[6 \lt 18\]\[\therefore\ a = 6\]

Mark scheme (d)

Scheme	Marks	AO
(i) Uses an appropriate trigonometric equation to find angle \(ADB\) or \(\dfrac{1}{2}ADB\) such as \(\arctan\dfrac{6}{18}\) or \(\arctan\dfrac{6}{8}\) OE PI by AWRT \(18.4^\circ\) or \(36.9^\circ\)	M1	3.1a
Obtains AWRT 0.644	A1	1.1b
	(2)
(ii) Uses \(l = r\theta\) with either \(r = 10\) or \(\theta = 2 \times\) their 0.644 OE Condone angle in degrees	M1	3.1a
Obtains AWFW [12.8,12.9]	A1	1.1b
	(2)
	(9 marks)

(d)(i)

\[\tan\frac{\theta}{2} = \frac{6}{18}\]\[\theta = 0.644\]

(d)(ii)

\[l = 10 \times (2 \times 0.644)\]\[= 12.9\]