Circles & Coordinate Geometry

Revision SessionAQA Core 1 Maths

Wednesday, March 11th

Starter

With a partner, list the following formulae:

• equation of a line

• gradient

• midpoint

• distance

• equation of a circle (describe how to find the centre & radius from the equation)

SolutionsEquation of a line:

11 xxmyy or cmxy

Gradient

21

21

xx

yym

Midpoint

2

,2

2121 yyxxM

Distance

2212

21 yyxxd

Equation of a circle:

222 rbyax

Where the centre is at point ba,

and the radius is r

Example 1 – The basics of straight lines

Given the points A (-3, 7) and B (5, 1), find:

(a) the midpoint between points A and B

(b) the distance between points A and B

(c) the gradient of the line that passes through AB

(d) an equation of the line that passes through point A and B

(e) the gradient of the line parallel to AB

(f) the gradient of the line perpendicular to AB

Solutions(a) midpoint:

2

17,

2

53M

Given the points A (-3, 7) and B (5, 1), find:

(b) distance: 22 1753 d

4,1

22 68 3664

(c) gradient:53

17

m8

6

10

4

3

(d) equation of line AB:

34

37 xy 3

4

37 xy

(e) gradient of line parallel to AB: 4

3parallelm

Note: If two lines with gradients and are parallel, then 1m 2m .21 mm

(f) gradient of line perpendicular to AB: 3

4larperpendicum

Note: If two lines with gradients and are perpendicular then 1m 2m121 mm

12

1mm

Exam Question – June 2013

If point lies on the line

Solutions

If a point lies on a line, then it should satisfy its equation. Meaning, we can substitute in the given point to find p.

0543 yx 2, ppthen

05243 pp

05843 pp031 p

31 p3p

Solutions

Rearrange the given equation into the form

m represents the gradient in this form.

.cmxy

0543 yx

534 xy

4

5

4

3

x

y

4

5

4

3x

y4

3 m

Solutions

If AC is perpendicular to AB, then

4

3ABm

3

4 ACm

Substitute the point A and C and the gradient of AC into the gradient formula in order to find the y-coordinate, k.

kC ,5:

3

4ACm

2,1: A21

21

xx

yym

51

2

3

4

k

6

2

3

4 k

k 23

24

k 28

k 10

k10

Solutions

If two lines intersect, then we can find their solution (where they intersect) by solving for x and y simultaneously!

0543 yxOriginal

652 yxNew

543 yx

652 yx

To eliminate x, multiply the top by 2 and the bottom by -

3, then add vertically.

5432 yx

6523 yx

1086 yx

18156 yx

287 y

4y

Substitute into either above equations to find the x coordinate. 652 yx 6452 x

6202 x7x

Therefore, point D lies at .4,7

You try! – timed (10 mins)Exam Question – January 2013

Swap solutions & peer assess

Label the Parts of a Circlekey terms:

centre

circumference

radius

diameter

chord

normal line

tangent line

major arc

minor arc

Example 2 – The basics of circlesA circle with centre C has equation

(a) Express this equation in the form

where a, b, and k are integers.

(b) Find the centre C.

(c) Find the radius of the circle

(d) The point D has coordinates (-2, -1). Verify that point D lies on the circle.

(e) Sketch the circle.

036422 yxyx

kbyax 22

Solutions(a) Start with the given equation and use “completing the square” technique to rearrange into the form

036422 yxyx

.22 kbyax

364 22 yyxx

39342 22 yx

1632 22 yx

SolutionsIf the circle’s equation is

(b) Then centre lies at

(c) and the radius is

(d) To show that point D lies on the circle, simply substitute the coordinates into the original equation and show that it does, in fact, equal zero.

1632 22 yx

3,2 416

036422 yxyx 1,2 D

03162412 22

036814 00

SolutionsIf the circle’s equation is

then the sketch looks like

1632 22 yx

Exam Question – June 2013

You try! – timed (10 mins)

Solutions

(a) Start with the given equation and use “completing the square” technique to rearrange into the form

025141022 yxyx

.22 kbyax

251410 22 yyxx

25497255 22 yx

4975 22 yx

Solutions

(b)(i) then the centre lies at .7,5

4975 22 yxIf

(ii) the radius is .749

Solutions

4975 22 yx(c)(i) Sketch of To get full marks, your circle must

• be in the correct position

• cut the negative y-axis twice

• touch the x-axis at x=5

• 5 must be marked on x-axis OR centre must be marked (either ok)

(c)(ii) The coordinates of the point that lies on the circle and is the furthest away from the x-axis would lie here. This is point (5, -14) because the centre is at (5, -7) and we know the radius is 7 units long.

Thus, we need to describe the transformation which maps the circle with equation onto the circle C.

Solutions

4975 22 yx(d) Solution to part (a) is

491 22 yx

Let’s take a look at the sketches of these two circles….6

4

2

-2

-4

-6

-8

-10

-12

-10 -5 5 10

4975 22 yx

491 22 yxBe careful with the wording of this type question! We need write a vector that would map the green circle “ONTO” to the black circle.

Translated

76 6 units right

7 units down

With the remaining time, try the past exam

questions given in your hand out & ask questions

when needed.