Question 10:
--- 10(a) ---
10(a) | Equation of circle is ( x − 3 ) 2 + ( y − 5 ) 2 = r 2 and line is y = 2 x + k
So intersect when ( x − 3 ) 2 + ( 2 x + k − 5 ) 2 = r 2 | M1
 x 2 − 6 x + 9 + 4 x 2 + 4 ( k − 5 ) x + ( k − 5 ) 2 = r 2
 5 x 2 + ( − 6 + 4 k − 2 0 ) x + 9 + k 2 − 1 0 k + 2 5 − r 2 = 0 | dM1
 5 x 2 + ( 4 k − 2 6 ) x + k 2 − 1 0 k + 3 4 − r 2 = 0 * | A1*
(3)
(b) | Tangent to C b2 −4ac=0(4k−26)2 −45 ( k2 −10k+34−r2) =0 | M1
 1 6 k 2 − 2 0 8 k + 6 7 6 − 2 0 k 2 + 2 0 0 k − 6 8 0 + 2 0 r 2 = 0
5r2 =... | M1
 5 r 2 = k 2 + 2 k + 1 = ( k + 1 ) 2 | A1
(3)
(b)
Way 2 | 1 1
Gradient of BX is − so equation of BX is y − 5 = − ( x − 3 )
2 2
1  1 3 − 2 k 2 6 + k 
y − 5 = − ( x − 3 ) , y = 2 x + k  x = ..., y = ... ,
2 5 5 | M1
 1 3 − 2 k  2  2 6 + k  2
− 3 + − 5 = r 2
5 5 | dM1
 5 r 2 = k 2 + 2 k + 1 = ( k + 1 ) 2 | A1
(3)
(c) | Triangle AXB is right angled so A B 2 + r 2 = X A 2 = ( 3 − 0 ) 2 + ( 5 − k ) 2 | M1
A B 2 = 4 r 2 so A B 2 + r 2 = 5 r 2 | M1
 5 r 2 = 9 + ( 5 − k ) 2 | A1
 k 2 + 2 k + 1 = 9 + 2 5 − 1 0 k + k 2 | M1
12k =33k =... | dM1
11
k =
4 | A1
(6)
(12 marks)
Notes:
(a)
M1: Forms equation of circle and substitutes in equation of line.
The circle equation must be of the form (x3)2 +(y5)2 =r2
dM1: Expands both sets of brackets and collects terms in x2 and x.
A1*: Reaches the answer given with no errors seen.
Note that some candidates expand the brackets first before substitution e.g.:
( x − 3 ) 2 + ( y − 5 ) 2 = r 2  x 2 − 6 x + 9 + y 2 − 1 0 y + 2 5 = r 2  x 2 − 6 x + 9 + ( 2 x + k ) 2 − 1 0 ( 2 x + k ) + 2 5 = r 2
This implies the first M and then the second M will score when terms in x2 and x are collected.
Note about poor squaring e.g. ( x − 3 ) 2 = x 2 + 9 : The first M is available in both cases above but
the second M requires at least two x2 terms and at least two x terms from the expansions.
Note that it is acceptable to go from a completely correct full expansion to the printed answer e.g.
( x − 3 ) 2 + ( y − 5 ) 2 = r 2  x 2 − 6 x + 9 + y 2 − 1 0 y + 2 5 = r 2  x 2 − 6 x + 9 + ( 2 x + k ) 2 − 1 0 ( 2 x + k ) + 2 5 = r 2
 x 2 − 6 x + 9 + 4 x 2 + 4 k x + k 2 − 2 0 x − 1 0 k + 2 5 = r 2  5 x 2 + ( 4 k − 2 6 ) x + k 2 − 1 0 k + 3 4 − r 2 = 0
Scores full marks in (a)
(b)
M1: Uses the discriminant is zero to form an equation in k and r
dM1: Expands and rearranges to make αr2 the subject
A1: Correct answer
Way 2
M1: Attempts the equation of BX and solves simultaneously with l to find the coordinates of B
b 1 3 − 2 k
Alternatively uses x = − = at B and uses this to find y at B
2 a 5
dM1: Correct use of Pythagoras for BX and sets = r2
A1: Correct answer
(c)
M1: Attempts XA2 correctly in terms of k (the k may appear as y but must be replaced by k later) and
uses it in Pythagoras theorem for triangle AXB.
May be implied.
M1: Applies AB = 2r to get A B 2 + r 2 in terms of r. Condone with AB2 = 2r2 used.
A1: Correct equation.
M1: Substitutes the result from (b) and expands brackets.
dM1: Solves a linear equation in k. Depends on the previous M.
A1: Correct value.
b 1 3 − 2 k  1 3 − 2 k 
At B x = − = , y = 2 + k
2 a 5 5 | M1
13−2k  2 26−4k  2
AB2 = −0 + +k−k
 
 5   5  | M1 A1
254r2 =( 13−2k )2 +(26−4k)2 =... | M1
20(k+1)2 =( 13−2k )2 +(26−4k)2 =20k2 −260k+845
1 6 9
 k 2 + 2 k + 1 = k 2 − 1 3 k +  k = ...
4 | dM1
11
k =
4 | A1
(6)
Notes:
(c)
M1: Uses the result in (a) to find the x coordinate where the line and circle meet and then finds y
An alternative is to find the equation of BX as in (b) way 2 and solve with l to find x and y at B
(May have already found the coordinates of B in (a) but must be re-stated or used in (c) to score
this mark)
M1: Uses distance formula to find an expression in k for AB or AB2
A1: Correct expression for AB or AB2. Need not be simplified.
M1: Applies AB = 2r to the equation. Condone with AB2 = 2r2 used.
dM1: Substitutes the result from (b) and solves a linear equation in k. Depends on the previous M.
A1: Correct value.
Way 5 | 1+3 0−10
If AB is diameter centre must be midpoint of AB ie M  , 
 2 2  | M1
= ( 2 , − 5 ) | A1
Attempts 2 of:
m = − 6 + 1 0 = 1 7+3 −6−10 and midpoint is  ,  =(5,−8)
B C 7 − 3  2 2 
1
so perpendicular bisector is y+"8"=− (x−"5")
"1"
or
m = 7 − 1 = − 1 and midpoint is  7 + 1 , − 6  = ( 4 , − 3 )
A C − 6 − 0 2 2
1
so perpendicular bisector is y + " 3 " = − ( x − " 4 " )
" − 1 "
or
m = − 1 0 − 0 = − 5 and midpoint is  3 + 1 , − 1 0  = ( 2 , − 5 )
A B 3 − 1 2 2
1
so perpendicular bisector is y + " 5 " = − ( x − " 2 " )
" − 5 "
y + 8 = − ( x − 5 ) oe or y + 3 = x − 4 1 oe or y+5= (x−2) oe
5
And solves simultaneously:
E.g. y + 3 = x − 4 , y + 8 = 5 − x  5 − x − 5 = x − 4  x = 2 , y = − 5 | dM1
Hence centre of circle is (2,−5) | A1
E.g. Midpoint of AB is the centre of the circle so AB is a diameter
(or equivalent) | A1
(5)
Way 6 | Uses ( x − a ) 2 + ( y − b ) 2 = r 2
With (1, 0), (7, −6) and (3, −10)
To find (a, b) = … or r/r2 = … | M1
Centre (2, −5) or radius 2 6 | A1
E.g. Equation of AB is y=−5 ( x−1 ) and −5 ( 2−1 )=−5
or AB= (3−1)2 +(−10−0)2 = 104 =2 26
 1 + 3 0 − 1 0 
or midpoint of AB is , = ( 2 , − 5 )
2 2 | dM1A1
So centre is on AB or AB is twice the radius or midpoint is the centre
hence AB is a diameter of the circle. (or equivalent) | A1
(5)
M1: Uses all three points in circle equation to set up three equations in three unknowns to find
centre or radius.
A1: Correct centre or correct radius
dM1: Finds e.g. equation of AB, distance AB or midpoint of AB
A1: Correct equation of AB, distance AB or midpoint of AB
A1: Suitable explanation and conclusion given with no errors and all previous marks awarded.
(ii)
Way 7 | 1+3 0−10
If AB is diameter centre must be midpoint of AB ie M  , 
 2 2  | M1
= ( 2,−5 ) | A1
Circle centre C radius r is ( x − 7 ) 2 + ( y + 6 ) 2 = r 2
Circle centre B radius r is ( x − 3 ) 2 + ( y + 1 0 ) 2 = r 2
These intersect when ( x − 7 ) 2 + ( y + 6 ) 2 = ( x − 3 ) 2 + ( y + 1 0 ) 2
x+ y=−3
Circle centre A radius r is ( x−1 )2 + y2 =r2
( x−3 )2 +( y+10 )2 =( x−1 )2 + y2  x−5y=27 | dM1A1
Solves simultaneously:
x + y = − 3 , x − 5 y = 2 7  x = 2 , y = − 5
E.g. Midpoint of AB is the centre of the circle so AB is a diameter
(or equivalent) | A1
(5)
M1: Attempts midpoint of AB. If no method is shown accept one correct coordinate as evidence.
A1: Correct midpoint
dM1: Attempts equations of 2 circles with A, B or C as centre with radius r, repeats the process for
2 different circles and finds the intersection of both straight lines and solves simultaneously
A1: Correct coordinates of centre
A1: Suitable explanation and conclusion given with no errors and all previous marks awarded.
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