Question 10:
--- 10(a) ---
10(a) | x 2 + y 2 + 4 x − 3 0 y + 2 0 9 = 0
 ( x  2 ) 2 + ( y  1 5 ) 2 . . . = 0 | M1
(i) | Centre (−2, 15) | A1
(ii) | Radius 2 0 | A1
(3)
(b) | y = m x + 1  ( x + 2 ) 2 + ( m x + 1 − 1 5 ) 2 = 2 0
or
y = m x + 1  x 2 + ( m x + 1 ) 2 + 4 x − 3 0 ( m x + 1 ) + 2 0 9 = 0 | M1
x2 +m2x2 +4x−28mx+180=0
b 2 − 4 a c = 0  ( 4 − 2 8 m ) 2 − 4 ( 1 + m 2 )  1 8 0 = 0 | dM1
( 4 − 2 8 m ) 2 − 4 ( 1 + m 2 ) 1 8 0 = 0  1 6 − 2 2 4 m + 7 8 4 m 2 − 7 2 0 − 7 2 0 m 2 = 0
 2 m 2 − 7 m − 2 2 = 0 * | A1*
(3)
(c) | 1 1
2 m 2 − 7 m − 2 2 = 0  m = , − 2
2 | M1
1 1 1 2 5 1 2 7 1
m =  x 2 − 1 5 0 x + 1 8 0 = 0  x =  y =
2 4 5 5
or
m=−25x2 +60x+180=0x=−6 y=13 | M1
 1 2 7 1 
, o r ( − 6 , 1 3 ) oe
5 5 | A1
 1 2 7 1 
, a n d ( − 6 , 1 3 ) oe
5 5 | A1
(4)
Total 10
(b) Alternative: Perpendicular distance from a point to a line:
( − 2 , 1 5 ) , m x − y + 1 = 0 with d =
a x
0
+
a 2
b y
+
0
b
+
2
c
−2m−15+1
20 = 20 ( m2 +1 ) =( 2m+14 )2
m2 +1
 1 6 m 2 − 5 6 m − 1 7 6 = 0  2 m 2 − 7 m − 2 2 = 0 *
M1: Substitutes into a correct distance formula with their centre and the given L and sets
equal to their radius.
dM1: Squares, multiplies up to obtain  ( Am2 +B ) =(Cm+D)2 oe, where A, B, C and D
are non-zero.
A1*: Obtains the printed answer with no errors
(c)
M1: Solves the given quadratic equation by any valid means, including calculator, to
obtain at least one value for m. May be implied by their value(s).
One correct value only with no incorrect work is sufficient for M1
Condone x = … for this mark.
M1: Uses at least one of their values of m to attempt one position for P.
This must be a complete and correct method to find a position for P i.e. finds at
least one value for x and then the corresponding value for y correctly.
E.g. substitutes at least one value for x into their ( x + 2 ) 2 + ( m x + 1 − 1 5 ) 2 = 2 0 ,
PMT
solves
the resulting 3TQ by any method including a calculator and then uses the correct
y = mx + 1 to find the y value. Condone slips when e.g. simplifying their 3TQ
providing the complete method is correct.
A1: At least one correct point. Allow as a coordinate pair or as x = …, y = … and allow
equivalent values e.g. (2.4, 14.2) etc.
A1: Both points correct and no others. Allow as coordinate pairs or as x = …, y = …
(c) Alternative 1:
Finds the intersections of the possible tangents with the perpendiculars passing through the
centre of the circle:
 11 
M1:  m= , −2 As above
 2 
For m =
1 1
2
perpendicular is y − 1 5 = −
1
2
1
( x + 2 )

y = −
1
2
1
x +
1 6 1
1 1

2 161 11 12 71
− x+ = x+1 x= , y =
11 11 2 5 5
or
For m = − 2 perpendicular is y − 1 5 =
1
2
( x + 2 )

y =
1
2
x + 1 6

PMT
1
x+16=−2x+1 x=−6, y =13
2
M1: For a complete method by
• forming the equation of at least one of the perpendiculars, with the negative reciprocal
gradient and the coordinates of their centre correctly placed
• solving simultaneously with the corresponding tangent to find x or y
• finding the corresponding x or y coordinate
A1: At least one correct point. Allow as a coordinate pair or as x = …, y = … and allow
equivalent values e.g. (2.4, 14.2) etc.
A1: Both points correct and no others. Allow as coordinate pairs or as x = …, y = …
(c) Alternative 2:
( 1 5 − 1 ) 2 + 2 2 = a 2 + ( 2 a ) 2 + ( 2 0 ) 2
or
2
( 15−1 )2 +22 =b2 +   11 b   + ( 20 )2
 2 
Then
a2 =36a=−6−2a+1=13
or
b 2 =
1 4
2
4
5
 b =
1 2
5

1 1
2
b + 1 =
7
5
1
 11 

b, b+1
 2 
Score as:
M1: As above
M1: For a complete method by
• using Pythagoras correctly to find the distance or distance2 between (0, 1) and their
centre
• applying Pythagoras correctly with a general point on either tangent and their radius
to obtain an equation in one variable and solves
• finding the corresponding x or y coordinate
A1: At least one correct point. Allow as a coordinate pair or as x = …, y = … and allow
equivalent values e.g. (2.4, 14.2) etc.
A1: Both points correct and no others. Allow as coordinate pairs or as x = …, y = …
y =
1 1
2
x + 1
( a , − 2 a + 1 )
( − 2 , 1 5 )
(
0 , 1
)
y = − 2 x + 1
PMT
General diagram for reference:
y = − 2 x + 1
y =
1
2
x + 1 6
y = −
1
2
1
x +
1
1
6 1
1
 1 2
5
,
7
5
1 
( − 6 , 1 3 )
( − 2 , 1 5 )
y =
1 1
2
x + 1
PMT