Edexcel A-Level Mathematics AS P2.3 Coordinate Geometry in the X Y Plane Questions

Practise circle coordinate geometry by finding centres, radii, equations, sketches and constraints from points or parameters.

Exam points

find a circle equation from a centre and a point, or from endpoints of a diameter
read the centre and radius from completed-square circle form
use radius and centre position to decide quadrant or touching conditions

Covered syllabus points

P2.3.1 - Coordinate geometry of the circle

[Maximum number: 3]

Figure 2

Figure 2 shows a sketch of
- the circle C with centre X(4,-3)
- the line l with equation $y=\frac{5}{2} x-\frac{55}{2}$

Given that l is the tangent to C at the point N,

Hence find

an equation for C.

M1: Attempts radius or radius ${ }^{2}$ . Condone slips on lhs, for example r or $d=(\text { " } 9 \text { " }-4)^{2}+(\text { " }-5 \text { " }+3)^{2}$ You must see an attempt to take the difference for both components before each are squared.

M1: Correct form of equation E.g. $(x-4)^{2}+(y+3)^{2}=" r "{ }^{2}$
Look for $(x-4)^{2}+(y+3)^{2}=(\text { " } 9 \text { " }-4)^{2}+(\text { " }-5 \text { " }+3)^{2}$
Condone for this mark a slip on one of the components when finding their $r^{2}$
A1: $(x-4)^{2}+(y+3)^{2}=29$ or exact equivalent. The constant terms must be collected and not 25+4 ISW after a correct answer.

Alt (b)(ii)
Note that an alternative version is $x^{2}+y^{2}-8 x+6 y-4=0$ . It is possible to proceed using this alternative version of the formula for a circle

M1: States equation of circle is $x^{2}+y^{2}-8 x+6 y+c=0$
M1: Substitutes their N=(9,-5) into $x^{2}+y^{2}-8 x+6 y+c=0$ o.e. and finds " c "

A1: $x^{2}+y^{2}-8 x+6 y-4=0$ o.e.

Figure for Answer to Question 1 — Edexcel A-Level Mathematics AS

[Maximum number: 6]

In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.

Figure 3

Figure 3 shows
- the curve $C_{1}$ with equation $y=x^{3}-5 x^{2}+3 x+14$
- the circle $C_{2}$ with centre T

The point T is the minimum turning point of $C_{1}$
Using Figure 3 and calculus,

(a)

Find an equation of the circle $C_{2}$

The line l shown in Figure 3, is the tangent to circle $C_{2}$ at A

[ 3 ]

At $A \quad y=8$ $r^{2}=(2-" 3 ")^{2}+(" 8 "-" 5 ")^{2} \quad(=10)$ $(x-3)^{2}+(y-5)^{2}=10$
(3)

(b)

Show that an equation of l is

y=\frac{1}{3} x+\frac{22}{3}

The region R, shown shaded in Figure 3, is bounded by $C_{1}, l$ and the y-axis.

[ 3 ]

$\frac{" 8 "-" 5 "}{2-" 3 "}=\ldots(-3)$ $y-" 8 "=" \frac{1}{3} "(x-2)$ $y=\frac{1}{3} x+\frac{22}{3} \quad *$

A1*

(3)

[Maximum number: 6]

Figure 1

The points P(23,14), Q(15,-30) and R(-7,-26) lie on the circle C, as shown in Figure 1.

(a)

Hence, or otherwise, find

[ 3 ]

(i)

the centre of C,

[ 1 ]

(ii)

the radius of C.

Given that the point S lies on C such that the distance QS is greatest,

[ 2 ]

r=\sqrt{(23-8)^2+(14+6)^2}

r=\frac12\sqrt{(23+7)^2+(14+26)^2}

r=25

M1: fully correct method for the radius.
A1: correct radius.

(b)

find an equation of the tangent to C at S, giving your answer in the form ax+by+c=0, where a, b and c are integers to be found.

[ 3 ]

S=(1,18)\quad\text{or}\quad m_T=\frac{7}{24}

m_N=\frac{18+6}{1-8}\Rightarrow m_T=\frac{8-1}{18+6}

y-18=\frac{7}{24}(x-1)

7x-24y+425=0

B1: correct coordinates for S or correct tangent gradient.
M1: correct straight-line method.
A1: correct equation.

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (i)

Question (ii)

Question (b)