Circles and Parabolas, Worked Solutions

Ques­tion 1: Equa­tion of a cir­cle from its cen­tre and radius#

The prob­lem#

Find the equa­tion of the cir­cle with

(a) cen­tre $(1, -2)$ and radius $4$;

(b) cen­tre $(-a, -b)$ and radius $\sqrt{a^2 - b^2}$.

Under­stand­ing the prob­lem#

In both parts you are given the cen­tre and the radius, and you must write the equa­tion of the cir­cle. The answer is expected in the gen­eral (expanded) form $x^2 + y^2 + 2gx + 2fy + c = 0$. In part (b) the cen­tre and radius con­tain let­ters $a$ and $b$; treat them like fixed num­bers.

The idea#

A cir­cle with cen­tre $(h, k)$ and radius $r$ has the stan­dard equa­tion

$$(x - h)^2 + (y - k)^2 = r^2.$$

We sub­sti­tute the given cen­tre and radius, then expand the squares and col­lect the terms.

Step-by-step solu­tion#

Part (a)

Step 1. Write $h = 1$, $k = -2$, $r = 4$ in the stan­dard form. Note that $y - (-2) = y + 2$.

$$(x - 1)^2 + (y + 2)^2 = 4^2 = 16.$$

Step 2. Expand each square using $(p \pm q)^2 = p^2 \pm 2pq + q^2$.

$$x^2 - 2x + 1 + y^2 + 4y + 4 = 16.$$

Step 3. Bring every term to the left side and sim­plify the con­stants: $1 + 4 - 16 = -11$.

$$x^2 + y^2 - 2x + 4y - 11 = 0.$$

Part (b)

Step 1. Here $h = -a$, $k = -b$, and $r^2 = a^2 - b^2$. So $x - h = x + a$ and $y - k = y + b$.

$$(x + a)^2 + (y + b)^2 = a^2 - b^2.$$

Step 2. Expand the left side.

$$x^2 + 2ax + a^2 + y^2 + 2by + b^2 = a^2 - b^2.$$

Step 3. Move every­thing to the left. The $a^2$ terms can­cel, and $b^2 - (-b^2) = 2b^2$.

$$\begin{aligned} x^2 + y^2 + 2ax + 2by + a^2 + b^2 - a^2 + b^2 &= 0 \\ x^2 + y^2 + 2ax + 2by + 2b^2 &= 0. \end{aligned}$$

Check­ing the answer#

For (a), com­pare with $x^2 + y^2 + 2gx + 2fy + c = 0$: $2g = -2$, $2f = 4$, $c = -11$, so the cen­tre is $(-g, -f) = (1, -2)$ and the radius is $\sqrt{1 + 4 + 11} = \sqrt{16} = 4$. Cor­rect.

For (b): $g = a$, $f = b$, $c = 2b^2$, so the cen­tre is $(-a, -b)$ and $r^2 = a^2 + b^2 - 2b^2 = a^2 - b^2$. Cor­rect.

Answer#

(a) $x^2 + y^2 - 2x + 4y - 11 = 0$; (b) $x^2 + y^2 + 2ax + 2by + 2b^2 = 0$.

Com­mon mis­take to avoid#

With a neg­a­tive coor­di­nate in the cen­tre, the sign flips inside the bracket: cen­tre $(1, -2)$ gives $(y + 2)^2$, not $(y - 2)^2$.

Ques­tion 2: Cen­tre and radius from the gen­eral equa­tion#

The prob­lem#

Find the cen­tre and the radius of each cir­cle:

(a) $x^2 + y^2 + 4x - 6y = 3$;

(b) $2x^2 + 2y^2 - x = 0$.

Under­stand­ing the prob­lem#

Each equa­tion is in expanded form. You must recover the cen­tre $(h, k)$ and the radius $r$. In part (b), the coef­fi­cients of $x^2$ and $y^2$ are $2$, not $1$, that is the con­di­tion to watch.

The idea#

We com­plete the square in $x$ and in $y$ to turn the equa­tion into the stan­dard form $(x - h)^2 + (y - k)^2 = r^2$. (Equiv­a­lently, use cen­tre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$, but only after the $x^2$ and $y^2$ coef­fi­cients are $1$.)

Step-by-step solu­tion#

Part (a)

Step 1. Group the $x$ terms and the $y$ terms.

$$(x^2 + 4x) + (y^2 - 6y) = 3.$$

Step 2. Com­plete each square. Half of $4$ is $2$ and $2^2 = 4$; half of $-6$ is $-3$ and $(-3)^2 = 9$. Add $4$ and $9$ to both sides so that the equa­tion stays bal­anced.

$$(x^2 + 4x + 4) + (y^2 - 6y + 9) = 3 + 4 + 9.$$

Step 3. Write the brack­ets as per­fect squares.

$$(x + 2)^2 + (y - 3)^2 = 16 = 4^2.$$

Step 4. Read off the cen­tre and radius: $x + 2 = x - (-2)$, so $h = -2$; $k = 3$; $r = 4$.

Part (b)

Step 1. Divide the whole equa­tion by $2$ so that $x^2$ and $y^2$ have coef­fi­cient $1$.

$$\displaystyle x^2 + y^2 - \tfrac{1}{2}x = 0.$$

Step 2. Com­plete the square in $x$. Half of $\displaystyle -\tfrac{1}{2}$ is $\displaystyle -\tfrac{1}{4}$, and $\displaystyle \left(-\tfrac{1}{4}\right)^2 = \tfrac{1}{16}$. Add $\displaystyle \tfrac{1}{16}$ to both sides.

$$\displaystyle \left(x^2 - \tfrac{1}{2}x + \tfrac{1}{16}\right) + y^2 = \tfrac{1}{16}.$$

Step 3. Write it as a per­fect square.

$$\displaystyle \left(x - \tfrac{1}{4}\right)^2 + (y - 0)^2 = \left(\tfrac{1}{4}\right)^2.$$

Step 4. The cen­tre is $\displaystyle \left(\tfrac{1}{4}, 0\right)$ and the radius is $\displaystyle \tfrac{1}{4}$.

Check­ing the answer#

For (b), the cir­cle has cen­tre $\displaystyle \left(\tfrac14, 0\right)$ and radius $\displaystyle \tfrac14$, so it should pass through the ori­gin. Put $x = 0, y = 0$ in $2x^2 + 2y^2 - x = 0$: $0 = 0$. It does.

For (a), the point $(2, 3)$ is $4$ units to the right of the cen­tre $(-2, 3)$, so it should lie on the cir­cle: $4 + 9 + 8 - 18 = 3$. Cor­rect.

Answer#

(a) Cen­tre $(-2, 3)$, radius $4$; (b) cen­tre $\displaystyle \left(\tfrac{1}{4}, 0\right)$, radius $\displaystyle \tfrac{1}{4}$.

Com­mon mis­take to avoid#

In (b), read­ing $g$ and $f$ before divid­ing by $2$ gives a wrong cen­tre. Always make the coef­fi­cient of $x^2$ and $y^2$ equal to $1$ first.

Ques­tion 3: Cir­cle through two points with its cen­tre on a line#

The prob­lem#

Find the equa­tion of the cir­cle pass­ing through $(2, 3)$ and $(-1, 1)$ whose cen­tre lies on the line $x - 3y - 11 = 0$.

Under­stand­ing the prob­lem#

Given: two points on the cir­cle, and a line on which the cen­tre sits. Asked: the equa­tion of the cir­cle. The cen­tre $(h, k)$ and radius are unknown, so we need two equa­tions for $h$ and $k$.

The idea#

This is like Exam­ple 4 of the les­son. Every point on a cir­cle is at the same dis­tance from the cen­tre, so the cen­tre is equidis­tant from $(2, 3)$ and $(-1, 1)$. That gives one equa­tion in $h$ and $k$. The cen­tre lying on the line gives the sec­ond. Solve them, then find $r^2$.

Step-by-step solu­tion#

Step 1. Let the cen­tre be $(h, k)$. Equal dis­tances to the two points (we square both sides to avoid square roots):

$$(h - 2)^2 + (k - 3)^2 = (h + 1)^2 + (k - 1)^2.$$

Step 2. Expand both sides.

$$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 + 2h + 1 + k^2 - 2k + 1.$$

Step 3. The $h^2$ and $k^2$ terms can­cel. Col­lect the rest.

$$\begin{aligned} -4h - 6k + 13 &= 2h - 2k + 2 \\ -6h - 4k &= -11 \\ 6h + 4k &= 11. \end{aligned}$$

Step 4. The cen­tre is on the line $x - 3y - 11 = 0$, so

$$h - 3k = 11 \quad\Longrightarrow\quad h = 11 + 3k.$$

Step 5. Sub­sti­tute $h = 11 + 3k$ into $6h + 4k = 11$.

$$\displaystyle \begin{aligned} 6(11 + 3k) + 4k &= 11 \\ 66 + 18k + 4k &= 11 \\ 22k &= -55 \\ k &= -\tfrac{5}{2}. \end{aligned}$$

Then $\displaystyle h = 11 + 3\left(-\tfrac{5}{2}\right) = 11 - \tfrac{15}{2} = \tfrac{7}{2}$.

Step 6. Find $r^2$ as the squared dis­tance from the cen­tre $\displaystyle \left(\tfrac72, -\tfrac52\right)$ to $(2, 3)$.

$$\displaystyle r^2 = \left(\tfrac{7}{2} - 2\right)^2 + \left(-\tfrac{5}{2} - 3\right)^2 = \left(\tfrac{3}{2}\right)^2 + \left(-\tfrac{11}{2}\right)^2 = \tfrac{9}{4} + \tfrac{121}{4} = \tfrac{130}{4} = \tfrac{65}{2}.$$

Step 7. Write the stan­dard form and expand.

$$\displaystyle \left(x - \tfrac{7}{2}\right)^2 + \left(y + \tfrac{5}{2}\right)^2 = \tfrac{65}{2}$$

$$\displaystyle x^2 - 7x + \tfrac{49}{4} + y^2 + 5y + \tfrac{25}{4} - \tfrac{130}{4} = 0.$$

The con­stants give $\displaystyle \tfrac{49 + 25 - 130}{4} = \tfrac{-56}{4} = -14$, so

$$x^2 + y^2 - 7x + 5y - 14 = 0.$$

Check­ing the answer#

Put $(2, 3)$: $4 + 9 - 14 + 15 - 14 = 0$. Put $(-1, 1)$: $1 + 1 + 7 + 5 - 14 = 0$. Both points lie on the cir­cle. The cen­tre $\displaystyle \left(\tfrac72, -\tfrac52\right)$ sat­is­fies $\displaystyle \tfrac72 + \tfrac{15}{2} - 11 = 0$, so it is on the line.

Answer#

$x^2 + y^2 - 7x + 5y - 14 = 0$.

Ques­tion 4: Cir­cle of radius 5 with cen­tre on the x-axis#

The prob­lem#

Find the equa­tion of the cir­cle with radius $5$ whose cen­tre lies on the $x$-axis and which passes through the point $(2, 3)$.

Under­stand­ing the prob­lem#

A point on the $x$-axis has $y$-coor­di­nate $0$, so the cen­tre is $(h, 0)$ with $h$ unknown. The radius is known: $5$. The cir­cle must pass through $(2, 3)$. Watch out: there may be more than one such cir­cle.

The idea#

The dis­tance from the cen­tre $(h, 0)$ to the point $(2, 3)$ must equal the radius $5$. This gives an equa­tion in $h$.

Step-by-step solu­tion#

Step 1. Use the dis­tance for­mula, squared.

$$(h - 2)^2 + (0 - 3)^2 = 5^2.$$

Step 2. Sim­plify.

$$(h - 2)^2 + 9 = 25 \quad\Longrightarrow\quad (h - 2)^2 = 16.$$

Step 3. Take square roots, remem­ber­ing both signs.

$$h - 2 = 4 \quad\text{or}\quad h - 2 = -4 \quad\Longrightarrow\quad h = 6 \quad\text{or}\quad h = -2.$$

Step 4. Write each cir­cle.

$$(x - 6)^2 + y^2 = 25 \qquad\text{or}\qquad (x + 2)^2 + y^2 = 25.$$

In expanded form these are $x^2 + y^2 - 12x + 11 = 0$ and $x^2 + y^2 + 4x - 21 = 0$.

Check­ing the answer#

Point $(2, 3)$ in the first: $(2 - 6)^2 + 9 = 16 + 9 = 25$. In the sec­ond: $(2 + 2)^2 + 9 = 25$. Both work.

Answer#

$(x - 6)^2 + y^2 = 25$ or $(x + 2)^2 + y^2 = 25$.

Com­mon mis­take to avoid#

Tak­ing only $h - 2 = 4$ loses the sec­ond cir­cle. A square root always gives $\pm$.

Ques­tion 5: Posi­tion of a point rel­a­tive to a cir­cle#

The prob­lem#

Does the point $(-2.5, 3.5)$ lie inside, on or out­side the cir­cle $x^2 + y^2 = 25$?

Under­stand­ing the prob­lem#

The cir­cle $x^2 + y^2 = 25$ has cen­tre $(0, 0)$ and radius $5$. A point is inside if its dis­tance from the cen­tre is less than $5$, on the cir­cle if equal to $5$, and out­side if greater.

The idea#

Com­pare the square of the dis­tance, $x^2 + y^2$, with $r^2 = 25$. This avoids tak­ing a square root.

Step-by-step solu­tion#

Step 1. Com­pute $x^2 + y^2$ for the point.

$$(-2.5)^2 + (3.5)^2 = 6.25 + 12.25 = 18.5.$$

Step 2. Com­pare with $25$.

$$18.5 < 25.$$

So the dis­tance from the cen­tre is $\sqrt{18.5} \approx 4.30$, which is less than the radius $5$.

Check­ing the answer#

Both coor­di­nates have size under $4$, so the point is well within $5$ of the ori­gin in each direc­tion; being inside is rea­son­able.

Answer#

The point lies inside the cir­cle.

Ques­tion 6: Focus, direc­trix and latus rec­tum of three parabo­las#

The prob­lem#

Find the coor­di­nates of the focus, the equa­tion of the direc­trix and the length of the latus rec­tum for

(a) $y^2 = 8x$; (b) $x^2 = 6y$; (c) $y^2 = -16x$.

Under­stand­ing the prob­lem#

Each equa­tion is a stan­dard parabola with ver­tex at the ori­gin. You must first decide which of the four stan­dard forms it matches, then find $a$.

The idea#

Use the table in the les­son:

The latus rec­tum always has length $4a$ (with $a > 0$).

Step-by-step solu­tion#

Part (a)

Step 1. $y^2 = 8x$ has the form $y^2 = 4ax$ (opens right). Match: $4a = 8$, so $a = 2$.

Step 2. Focus $(a, 0) = (2, 0)$; direc­trix $x = -2$; latus rec­tum $4a = 8$.

Part (b)

Step 1. $x^2 = 6y$ has the form $x^2 = 4ay$ (opens up). Match: $4a = 6$, so $\displaystyle a = \tfrac{3}{2}$.

Step 2. Focus $\displaystyle \left(0, \tfrac32\right)$; direc­trix $\displaystyle y = -\tfrac32$; latus rec­tum $4a = 6$.

Part (c)

Step 1. $y^2 = -16x$ has the form $y^2 = -4ax$ (opens left). Match: $4a = 16$, so $a = 4$.

Step 2. Focus $(-4, 0)$; direc­trix $x = 4$; latus rec­tum $16$.

Check­ing the answer#

In each case the focus and direc­trix are on oppo­site sides of the ver­tex, at equal dis­tance $a$, and the focus is on the side towards which the parabola opens. For (a), the point $(2, 4)$ lies on $y^2 = 8x$; its dis­tance to the focus $(2, 0)$ is $4$ and its dis­tance to the line $x = -2$ is $4$. Equal, as the def­i­n­i­tion requires.

Answer#

(a) Focus $(2, 0)$, direc­trix $x = -2$, latus rec­tum $8$; (b) focus $\displaystyle \left(0, \tfrac32\right)$, direc­trix $\displaystyle y = -\tfrac32$, latus rec­tum $6$; (c) focus $(-4, 0)$, direc­trix $x = 4$, latus rec­tum $16$.

Com­mon mis­take to avoid#

The latus rec­tum is $4a$, not $a$ or $2a$, and it is always a pos­i­tive length even when the parabola opens left or down.

Ques­tion 7: Parabola from focus and direc­trix#

The prob­lem#

Find the equa­tion of the parabola with focus $(0, -4)$ and direc­trix $y = 4$.

Under­stand­ing the prob­lem#

The focus is on the $y$-axis, below the ori­gin; the direc­trix is the hor­i­zon­tal line $4$ units above the ori­gin. The ver­tex is mid­way between them, at the ori­gin. The parabola opens away from the direc­trix, towards the focus, that is, down­wards.

The idea#

A parabola with ver­tex at the ori­gin open­ing down is $x^2 = -4ay$, with focus $(0, -a)$ and direc­trix $y = a$. Match $a$.

Step-by-step solu­tion#

Step 1. Com­pare the focus $(0, -4)$ with $(0, -a)$: $a = 4$. The direc­trix $y = a = 4$ agrees.

Step 2. Sub­sti­tute into $x^2 = -4ay$.

$$x^2 = -4(4)y = -16y.$$

Step 3 (from the def­i­n­i­tion, as a sec­ond route). For a point $(x, y)$, dis­tance to focus $=$ dis­tance to direc­trix:

$$\begin{aligned} x^2 + (y + 4)^2 &= (y - 4)^2 \\ x^2 + y^2 + 8y + 16 &= y^2 - 8y + 16 \\ x^2 &= -16y. \end{aligned}$$

Check­ing the answer#

The point $(8, -4)$ lies on $x^2 = -16y$ since $64 = 64$. Its dis­tance to the focus $(0, -4)$ is $8$; its dis­tance to the line $y = 4$ is $|-4 - 4| = 8$. Equal.

Answer#

$x^2 = -16y$.

Ques­tion 8: Parabola with a given axis through a point#

The prob­lem#

Find the equa­tion of the parabola with ver­tex $(0, 0)$, axis along the $x$-axis, pass­ing through $(3, -6)$.

Under­stand­ing the prob­lem#

Axis along the $x$-axis means the equa­tion is $y^2 = 4ax$ or $y^2 = -4ax$. The point $(3, -6)$ has a pos­i­tive $x$-coor­di­nate, so the parabola opens to the right.

The idea#

Take $y^2 = 4ax$ and use the given point to find $a$ (as in Exam­ple 7).

Step-by-step solu­tion#

Step 1. Sub­sti­tute $x = 3$, $y = -6$ into $y^2 = 4ax$.

$$(-6)^2 = 4a(3) \quad\Longrightarrow\quad 36 = 12a \quad\Longrightarrow\quad a = 3.$$

Step 2. Write the equa­tion.

$$y^2 = 4(3)x = 12x.$$

Check­ing the answer#

$(-6)^2 = 36$ and $12 \times 3 = 36$. The point lies on it, and $a = 3 > 0$ con­firms it opens right.

Answer#

$y^2 = 12x$.

Com­mon mis­take to avoid#

The neg­a­tive $y$-coor­di­nate does not make the parabola open left, only the sign of $x$ decides that, because $y$ is squared.

Ques­tion 9: Tri­an­gle formed by the ver­tex and the latus rec­tum#

The prob­lem#

Find the area of the tri­an­gle formed by the ver­tex of the parabola $x^2 = 12y$ and the ends of its latus rec­tum.

Under­stand­ing the prob­lem#

We need three points: the ver­tex, and the two end­points of the latus rec­tum (the chord through the focus per­pen­dic­u­lar to the axis). Then we find the area of the tri­an­gle they make.

The idea#

For $x^2 = 4ay$ the focus is $(0, a)$ and the latus rec­tum lies on the line $y = a$. Its ends are found by putting $y = a$ into the equa­tion. The tri­an­gle then has a hor­i­zon­tal base and a height equal to $a$.

Step-by-step solu­tion#

Step 1. Match $x^2 = 12y$ with $x^2 = 4ay$: $4a = 12$, so $a = 3$. Ver­tex $(0, 0)$, focus $(0, 3)$.

Step 2. Put $y = 3$ into the equa­tion to find the ends of the latus rec­tum.

$$x^2 = 12 \times 3 = 36 \quad\Longrightarrow\quad x = \pm 6.$$

The ends are $(6, 3)$ and $(-6, 3)$.

Step 3. The base joins $(-6, 3)$ and $(6, 3)$, so its length is $12$ (this is the latus rec­tum, $4a = 12$). The height is the dis­tance from the ver­tex $(0, 0)$ to the line $y = 3$, which is $3$.

Step 4. Area of the tri­an­gle:

$$\displaystyle \text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 12 \times 3 = 18.$$

Check­ing the answer#

With the coor­di­nate for­mula: $\displaystyle \tfrac12\,|0(3 - 3) + 6(3 - 0) + (-6)(0 - 3)| = \tfrac12\,|18 + 18| = 18$. Same result.

Answer#

The area is $18$ square units.

Ques­tion 10: Width of a par­a­bolic arch#

The prob­lem#

An arch is in the form of a parabola with its axis ver­ti­cal. The arch is $8$ m wide at the base and $6$ m high. How wide is it $2$ m below the top?

Under­stand­ing the prob­lem#

Draw the arch: it rises from the ground to a high­est point $6$ m up, and at ground level the two feet are $8$ m apart. We want the width of the arch at the height that is $2$ m below the top (that is, $4$ m above the ground).

The idea#

Choose axes to make the equa­tion sim­ple, as in Exam­ple 8: put the ver­tex at the top of the arch with the $y$-axis ver­ti­cal. The arch opens down­wards, so its equa­tion is $x^2 = -4ay$. The base is then $6$ m below the ver­tex.

Step-by-step solu­tion#

Step 1. Set up coor­di­nates. With the ver­tex at the ori­gin, the feet of the arch are at $y = -6$, and they are $4$ m either side of the axis (half of $8$). So the point $(4, -6)$ lies on the arch.

Step 2. Find $a$ by sub­sti­tut­ing $(4, -6)$ into $x^2 = -4ay$.

$$\displaystyle 16 = -4a(-6) = 24a \quad\Longrightarrow\quad a = \tfrac{16}{24} = \tfrac{2}{3}.$$

So the arch is $\displaystyle x^2 = -\tfrac{8}{3}y$.

Step 3. "$2$ m below the top" means $y = -2$. Sub­sti­tute.

$$\displaystyle x^2 = -\tfrac{8}{3}(-2) = \tfrac{16}{3} \quad\Longrightarrow\quad x = \pm\sqrt{\tfrac{16}{3}} = \pm\tfrac{4}{\sqrt{3}}.$$

Step 4. The width is the dis­tance between the two points, $\displaystyle 2 \times \tfrac{4}{\sqrt{3}}$.

$$\displaystyle \text{Width} = \tfrac{8}{\sqrt{3}} = \tfrac{8\sqrt{3}}{3} \approx 4.62 \text{ m}.$$

Check­ing the answer#

The width near the top must be less than the $8$ m base, and it is ($4.62 < 8$). Also, width is pro­por­tional to $\sqrt{\text{depth below vertex}}$: at depth $6$ the width is $8$, so at depth $2$ it is $8\sqrt{2/6} = 8/\sqrt{3}$. Same.

Answer#

The arch is $\displaystyle \tfrac{8}{\sqrt{3}} = \tfrac{8\sqrt3}{3} \approx 4.62$ m wide at $2$ m below the top.

Com­mon mis­take to avoid#

Using $y = 2$ (mea­sured from the ground) in an equa­tion whose ver­tex is at the top. Always mea­sure $y$ from wher­ever you placed the ori­gin.