The Maths Tailor

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.

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

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.

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

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.

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

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.

Find the complete set of values of

p

for which the equation

x^{2} - 2 p x + y^{2} - 6 y - p^{2} + 8 p + 9 = 0

describes a circle in the

x y

-plane.

(A)

p < - \frac{9}{4}

(B)

0 < p < 4

(C)

- 1 < p < 9

(D)

p < 0

p > 4

(E)

p < - 1

p > 9

(F) all real values of

p

Complete the square in both

x

and

y

(x - p)^{2} - p^{2} + (y - 3)^{2} - 9 - p^{2} + 8 p + 9 = 0

, which simplifies to

(x - p)^{2} + (y - 3)^{2} = 2 p^{2} - 8 p

. This is a circle if and only if the right-hand side is strictly positive:

2 p^{2} - 8 p > 0

, i.e.

p (p - 4) > 0

. This inequality holds when

p < 0

p > 4

. Hence the answer is D.

The point

P

has coordinates

(p, q)

, and the equation of a circle is

x^{2} + 2 f x + y^{2} + 2 g y + h = 0

where

f

g

h

p

and

q

are all real constants. Let

L

be the distance between the centre of the circle and the point

P

. Which one of the following is sufficient on its own to be able to calculate

L

?

(A) the values of

f

g

and

h

(B) the values of

f

g

p

and

q

f

h

p

and

q

(D) the values of

g

h

p

and

q

(E) none of the options A-D is sufficient on its own

Completing the square:

(x + f)^{2} + (y + g)^{2} = f^{2} + g^{2} - h

, so the centre is

(- f, - g)

. The distance from the centre to

P = (p, q)

L = (p + f)^{2} + (q + g)^{2}

. This requires

f

g

p

and

q

, but not

h

. Hence option B (the values of

f

g

p

and

q

) is sufficient. Hence the answer is B.

The radius of the circle

2 x^{2} + 2 y^{2} - 8 x + 12 y + 15 = 0

is

(A)

\frac{5}{2}

(B)

\frac{11}{2}

(C)

\frac{41}{2}

(D)

37

(E)

67

Divide through by 2:

x^{2} + y^{2} - 4 x + 6 y + \frac{15}{2} = 0

. Complete the square:

(x - 2)^{2} - 4 + (y + 3)^{2} - 9 + \frac{15}{2} = 0

, so

(x - 2)^{2} + (y + 3)^{2} = 4 + 9 - \frac{15}{2} = 13 - \frac{15}{2} = \frac{11}{2}

. The radius is

\frac{11}{2}

. Hence the answer is B.