CAIE A-Level Mathematics AS 1.3.4 Lines and Circles Questions

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