Coordinate Geometry & Circles

7 methods · 25 questions

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The instant an equation has $x^{2}$ and $y^{2}$ with linear terms, complete the square in each variable to reach $(x - a)^{2} + (y - b)^{2} = r^{2}$ , then read off centre $(a, b)$ and radius $r$ . For a sideways parabola $x = y^{2} + \dots$ , complete the square in $y$ to expose the vertex. If the leading coefficient is not 1, divide through first.

Trigger: a circle in general form $x^{2} + y^{2} + D x + E y + F = 0$ , or any second-degree equation where the centre/vertex is not yet visible.

Instances:
(i) divide by the common factor, then complete the square to get the radius;
(ii) complete the square in both $x$ and $y$ to expose centre and $r^{2}$ ;
(iii) complete the square in $y$ for $x = y^{2} + b y + c$ to find the parabola's vertex.

Linked questions (3)

TMUA 2017 Paper 1 · Q9

A circle has equation

x^{2} + y^{2} - 18 x - 22 y + 178 = 0

. A regular hexagon is drawn inside this circle so that the vertices of the hexagon touch the circle. What is the area of the hexagon?

A $6$
B $63$
C $18$
D $183$
E $36$
F $363$
G $48$
H $483$

Worked solution

Answer:F

Complete the square to find the radius:

(x - 9)^{2} - 81 + (y - 11)^{2} - 121 + 178 = 0 ⟹ (x - 9)^{2} + (y - 11)^{2} = 24.

r^{2} = 24

. A regular hexagon inscribed in a circle of radius

r

is composed of 6 equilateral triangles, each with side length

r

. The area of one equilateral triangle with side

r

\frac{3}{4} r^{2}

. Total area of hexagon:

6 \times \frac{3}{4} r^{2} = \frac{3 3}{2} \times 24 = 363 .

Hence the answer is F.

TMUA 2021 Paper 1 · Q18

The curve with equation

x = y^{2} - 6 y + 11

is rotated

9 0^{\circ}

clockwise about the point

P

to give the curve

C

P

has

x

-coordinate

- 2

and

y

-coordinate

3

. What is the equation of

C

A $y = - x^{2} - 4 x - 3$
B $y = - x^{2} - 4 x - 5$
C $y = - x^{2} - 6 x - 7$
D $y = - x^{2} - 6 x - 11$
E $y = x^{2} - 4 x + 5$
F $y = x^{2} + 4 x + 3$
G $y = x^{2} - 6 x + 11$
H $y = x^{2} + 6 x + 7$

Worked solution

Answer:B

Complete the square:

x = (y - 3)^{2} + 2

, so the parabola has vertex at

(2, 3)

, which is 4 units to the right of

P = (- 2, 3)

. A

9 0^{\circ}

clockwise rotation maps the vector

(4, 0)

(from

P

to vertex) to

(0, - 4)

, placing the new vertex at

(- 2, 3 - 4) = (- 2, - 1)

. The original parabola opens rightward (increasing

x

from vertex); after

9 0^{\circ}

clockwise rotation it opens downward. The rotated parabola has vertex

(- 2, - 1)

and opens downward, with equation

y - (- 1) = - (x - (- 2))^{2}

, i.e.

y = - (x + 2)^{2} - 1 = - x^{2} - 4 x - 4 - 1 = - x^{2} - 4 x - 5

. Hence the answer is B.

TMUA 2021 Paper 2 · Q15

A circle has equation

x^{2} + a x + y^{2} + b y + c = 0

where

a

b

and

c

are non-zero real constants. Which one of the following is a necessary and sufficient condition for the circle to be tangent to the

y

-axis?

A $a^{2} = 4 c$
B $b^{2} = 4 c$
C $\frac{a}{2} = \frac{a ^{2} + b ^{2}}{4} - c$
D $\frac{b}{2} = \frac{a ^{2} + b ^{2}}{4} - c$
E $- \frac{a}{2} = \frac{a ^{2} + b ^{2}}{4} - c$
F $- \frac{b}{2} = \frac{a ^{2} + b ^{2}}{4} - c$

Worked solution

Answer:B

Complete the square:

(x + \frac{a}{2})^{2} + (y + \frac{b}{2})^{2} = \frac{a ^{2}}{4} + \frac{b ^{2}}{4} - c

. The circle has centre

(- \frac{a}{2}, - \frac{b}{2})

and radius

r = \frac{a ^{2} + b ^{2}}{4} - c

. The circle is tangent to the

y

-axis if and only if the radius equals the distance of the centre from the

y

-axis, i.e.

r = - \frac{a}{2} = \frac{∣ a ∣}{2}

. Squaring:

\frac{a ^{2} + b ^{2}}{4} - c = \frac{a ^{2}}{4}

, which simplifies to

\frac{b ^{2}}{4} = c

, i.e.

b^{2} = 4 c

. (Both sides are non-negative so squaring is an equivalence.) Hence the answer is B.