OCR MEI AS Paper 2 2018 June — Question 8 7 marks

Exam BoardOCR MEI
ModuleAS Paper 2 (AS Paper 2)
Year2018
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeDistance from centre to line
DifficultyStandard +0.3 This is a standard coordinate geometry question requiring students to find the circle's radius, the line's equation, and verify tangency by showing the perpendicular distance from centre to line equals the radius. While it involves multiple steps (distance formula, line equation, perpendicular distance formula), these are all routine AS-level techniques with no novel insight required, making it slightly easier than average.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle

In this question you must show detailed reasoning. The centre of a circle C is at the point \((-1, 3)\) and C passes through the point \((1, -1)\). The straight line L passes through the points \((1, 9)\) and \((4, 3)\). Show that L is a tangent to C. [7]

Question 8:
AnswerMarks
8(x1)2 (y3)2 r2
r2 (11)2 (13)2
L: m = ‒ 2
y = ‒ 2x + 11 oe
substitution of their y = ‒ 2x + 11 in their
(x1)2 (y3)2 20
x2 ‒ 6x + 9 = 0 oe
(x ‒ 3)2 = 0 so repeated root
AnswerMarks
Hence line touches the curve and is a tangentM1
M1
B1
B1
M1
A1
E1
AnswerMarks
[7]2.1
1.1
1.1
1.1
1.1
1.1
AnswerMarks
2.4Left side correct and =
Or find L first (B1B1), then find
equation of line perp to L through
(‒1, 3) (M1M1) then substitute
(M1), solve (A1) then check (E1).
soi
AnswerMarks
or (‒ 6)2 ‒ 4×1×9 = 0or line through centre which is
perpendicular to L has
equation
y ‒ 3 = ½(x ‒ ‒ 1)
meets L at (3, 5)
(3 + 1)2 + (5 ‒ 3)2 = r2 =20 so
lines meet at circumference of
circle at
right angles so L is a tangent
Question 8:
8 | (x1)2 (y3)2 r2
r2 (11)2 (13)2
L: m = ‒ 2
y = ‒ 2x + 11 oe
substitution of their y = ‒ 2x + 11 in their
(x1)2 (y3)2 20
x2 ‒ 6x + 9 = 0 oe
(x ‒ 3)2 = 0 so repeated root
Hence line touches the curve and is a tangent | M1
M1
B1
B1
M1
A1
E1
[7] | 2.1
1.1
1.1
1.1
1.1
1.1
2.4 | Left side correct and =
Or find L first (B1B1), then find
equation of line perp to L through
(‒1, 3) (M1M1) then substitute
(M1), solve (A1) then check (E1).
soi
or (‒ 6)2 ‒ 4×1×9 = 0 | or line through centre which is
perpendicular to L has
equation
y ‒ 3 = ½(x ‒ ‒ 1)
meets L at (3, 5)
(3 + 1)2 + (5 ‒ 3)2 = r2 =20 so
lines meet at circumference of
circle at
right angles so L is a tangent
In this question you must show detailed reasoning.

The centre of a circle C is at the point $(-1, 3)$ and C passes through the point $(1, -1)$. The straight line L passes through the points $(1, 9)$ and $(4, 3)$. Show that L is a tangent to C. [7]

\hfill \mbox{\textit{OCR MEI AS Paper 2 2018 Q8 [7]}}