Question 6:
--- 6(a) ---
6(a) | Begins a valid method to find the
coordinates
Uses gradient of L to find gradient
of perpendicular radius
Or
Forms equation of circle with
unknown radius and solves
simultaneously with equation of L
Or differentiates equation of circle
implicitly | 3.1a | M1 | 5y+12x =298
−12 298
y = x+
5 5
5
y−9= ( x−7 )
12
12y−5x =73
x =19
y =14
( )
19,14
Uses (7, 9) to find the equation of
the radius
Or
Uses (7, 9) correctly in their
equation of circle
Or
−12
Uses after their implicit
5
differentiation | 1.1a | M1
Obtains 12y−5x=73OE
Or
Correctly eliminates a variable to
obtain a quadratic in x or y for
example
obtain a quadratic in x or y
298−12x  2
( x−7 )2 + −9 =k

 5 
253−12x 2
⇒( x−7 )2 + =k
 
 5 
298−5y  2
−7 +( y−9 )2 =k

 12 
214−5y 2
⇒ +( y−9 )2 =k
 
 12  | 1.1b | A1
Equates discriminant to zero
PI By correct answer
or
Solves their simultaneous
equations of tangent and radius
PI by correct answer | 3.1a | M1
Obtains correct values for x and y
( )
19,14 | 1.1b | A1
Subtotal | 5
--- 6(b) ---
6(b) | Obtains
(x−7)2+(y−9)2 =r2
r2
PI if is replaced with correct
value using their point from (a) | 1.1a | M1 | (x−7)2+(y−9)2 =r2
( 19−7 )2 +( 14−9 )2 =r2
122+52 =169
(x−7)2 +(y−9)2 =169
Uses their point to find radius or
radius squared.
Must have obtained a point in part
(a) | 1.1a | M1
Obtains correct equation of circle
CSO
ACF | 1.1b | A1
Subtotal | 3
Question Total | 8
Q | Marking instructions | AO | Mark | Typical solution