CAIE A-Level Mathematics AS 1.3 Coordinate Geometry Questions

Gradient of $A B=\frac{2-(-1)}{5-2}$
Expect 1, must be from $\Delta y / \Delta x$.

Equation of A B is y-2=1(x-5) or y+1=1(x-2)
OE. Expect y=x-3.

$k(5-2 x)^{\frac{1}{2}}=-3$
Equating their $\frac{\mathrm{d} y}{\mathrm{~d} x}$ of the form $k(5-2 x)^{\frac{1}{2}}$ to -3 .

[B is ](2,6)
Gradient $A B=m_{1}=\frac{32-6}{-2-2}$,gradient of perpendicular $=-\frac{1}{m_{1}}=\frac{4}{26}$

M1*

For A, y must be 32 .
Clear use of $\frac{\text { difference in y co-ordinates }}{\text { difference in x co-ordinates }}$ for points
A and B, condone inconsistent order, and using
$m_{1} m_{2}=-1$.
If incorrect values or another complete method used, then working must be clear.

Mid point is $\left(\frac{2-2}{2}, \frac{6+32}{2}\right)=(0,19)$

M1*

Finding the midpoint of A B using A and B. If
incorrect values used then all working must be clear.
For A, y must be 32 .

$y-19=\frac{2}{13}(x-0)$
Finding the equation of the perpendicular bisector using their midpoint and their perpendicular
gradient.

2 x-13 y+247=0 or $\pm$ integer multiples of this.

State or imply $(x+4)^{2}+(y-5)^{2}=20$
If $x^{2}+y^{2}-2 g x-2 f y+c=0$ is used correctly with
(-g,-f)=(-4,5) and $c=g^{2}+f^{2}-r^{2}$ then M1.

Obtain $x^{2}+y^{2}+8 x-10 y+21=0$
A1 if above method used.

Substitute x=0 and attempt solution of 3-term quadratic equation in y
If y=0 used can score a maximum of M0 A0 B1
M1 A0 A1FT DM1 A0, i.e. 4/8.

-5 and 3
B1 SC if no working to solve the quadratic.

State or imply centre of circle is (3,-1)
Condone errors which don't affect finding centre.
May be implied by the correct final y coordinate.

Attempt gradient of A C or B C

*M1

$-\frac{4}{3}$ or $\frac{4}{3}$
State or imply gradient of tangent is $\frac{3}{4}$ or $-\frac{3}{4}$

A1FT

Following their gradient of radius.
Only FT when previous 2 marks are M1 A0.

Either solve simultaneous equations (of 2 tangent equations) to find x - coordinate
Or Substitute y-value of centre into either tangent equation
$x=-\frac{16}{3}, y=-1$
Alternative Method 1: for the 4th and 5th marks

Rearrange and differentiate the circle equation or differentiate implicitly
Replaces the second M1.

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3-x}{y+1}$ or $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3-x}{\left(25-(x-3)^{2}\right)^{\frac{1}{2}}}$
Replaces the second A1.

8

Alternative Method 2: for the last 5 marks

$\widehat{A C P}=\widehat{M A P}=\tan ^{-1} \frac{4}{3}$ or identifying similar triangles P M A and A M C
C is the circle centre, P is intersection of the two
tangents, M is intersection of P C and the y-axis.

$\tan M A P=\frac{P M}{4}, \frac{4}{3}=\frac{P M}{4}, P M=\frac{16}{3}$ or use of similar triangles
P is $\left(\frac{-16}{3},-1\right)$
Alternative Method 3: for the last 5 marks

Pythagoras on triangle $P A C, P C^{2}=P A^{2}+A C^{2}$,
Identifies the required 3 sides and sets up
formula.

$P C^{2}=(P M+3)^{2}, P A^{2}=P M^{2}+4^{2}, A C=$ radius =5
Finds each side with two in terms of P M OE.

$(P M+3)^{2}=P M^{2}+4^{2}+5^{2}$ leads to $6 P M=32, P M=\frac{16}{3}$
Sets up and solves equation.

P is $\left(\frac{-16}{3},-1\right)$