Circles with Unknowns and Axes

Equation containing k: centre and radius in terms of k; substitute a point to get a quadratic in k

In the exam for Edexcel International AS Maths · 4 real past-paper questions below

Circles with a letter k

You can already read the centre and radius of a circle. AS questions often put an unknown constant, $k$ , into the equation. The method is the same: the centre and $r^{2}$ just come out in terms of $k$ .

Worked example

1
A circle has equation $(x - 2 k)^{2} + (y + k)^{2} = k + 8$ . Write down its centre and radius in terms of $k$ .
2
$Centre (2 k, - k)$ — Flip the signs inside the brackets, as usual.
3
$r = k + 8$ — Square root of the right-hand side.

Now suppose the circle passes through a known point. Its x and y must make the equation true, so put them in. You get an equation in $k$ , usually a quadratic.

Worked example

1
The circle passes through $(3, 1)$ . Find the possible values of $k$ .
2
$(3 - 2 k)^{2} + (1 + k)^{2} = k + 8$ — Put $x = 3$ , $y = 1$ into the equation.
3
$9 - 12 k + 4 k^{2} + 1 + 2 k + k^{2} = k + 8$ — Multiply out both brackets.
4
$5 k^{2} - 11 k + 2 = 0$ — Collect everything on one side.
5
$(5 k - 1) (k - 2) = 0$ — Factorise.
6
$k = \frac{1}{5} or k = 2$ — Both give $k + 8 > 0$ , so both are real circles.

Common mistake

Not squaring the whole bracket. $(3 - 2 k)^{2}$ is $9 - 12 k + 4 k^{2}$ , not $9 + 4 k^{2}$ .

Rule

Sometimes the question prints the quadratic: “show that $5 k^{2} - 11 k + 2 = 0$ ”. Then the last mark needs every line shown, with no errors. Do not jump to the printed line.

A circle must have a real radius, so $r^{2}$ must be more than 0. That gives the possible values of $k$ .

Worked example

1
$x^{2} + y^{2} - 8 x + 2 y = k$ is a circle. State the possible range of values of $k$ .
2
$(x - 4)^{2} - 16 + (y + 1)^{2} - 1 = k$ — Complete the square on x and on y.
3
$(x - 4)^{2} + (y + 1)^{2} = k + 17$ — Move the numbers to the right.
4
$k + 17 > 0$ — The right-hand side is $r^{2}$ , so it must be positive.
5
$k > - 17$ — The range.

Common mistake

Writing $k = - 17$ . The mark needs an inequality, $k > - 17$ .

Key idea

A letter in the equation changes nothing. Find the centre and $r^{2}$ in terms of $k$ , then use what the question tells you.

Touching an axis, highest point

Picture a coin standing on a table. Its centre is exactly one radius above the table. A circle that touches the x-axis is the same.

Rule

A circle with centre $(a, b)$ touches the x-axis when $r = ∣ b ∣$ . It touches the y-axis when $r = ∣ a ∣$ .

Memory trick

The axis is the floor. The radius is how high the centre sits off the floor. For the x-axis, that height is the y-coordinate, ignoring its sign.

Worked example

1
The circle $x^{2} + y^{2} + 10 x - 6 y + k = 0$ touches the y-axis. Find $k$ .
2
$(x + 5)^{2} + (y - 3)^{2} = 34 - k$ — Complete the square: $25 + 9 - k = 34 - k$ .
3
$Centre (- 5, 3)$ — Read it off.
4
$r = 5$ — The centre is 5 units from the y-axis.
5
$34 - k = 25$ — $r^{2} = 5^{2}$ .
6
$k = 9$ — Solve.

Example

Another way: touching means meeting the axis at one point. Put $x = 0$ to get $y^{2} - 6 y + k = 0$ . One root means the discriminant is 0: $36 - 4 k = 0$ , so $k = 9$ . Both ways are accepted.

The same idea finds the top, bottom and sides of a circle. Move straight up, down, left or right from the centre by one radius.

Formula

top (a, b + r) bottom (a, b - r)

$(a, b)$ — the centre.
$r$ — the radius. For the far left and right, use $(a \pm r, b)$ .

Worked example

1
Find the exact point on $(x - 2)^{2} + (y - 7)^{2} = 20$ that is nearest the x-axis.
2
$Centre (2, 7), r = 20 = 25$ — Read off and simplify.
3
$7 - 25 > 0$ — About 2.5, so the whole circle is above the x-axis.
4
$(2, 7 - 25)$ — The bottom of the circle is nearest the x-axis.

Common mistake

Giving $y = 2.53$ . “Exact” means keep the surd. Give both coordinates.

Key idea

A Ferris wheel works the same way. The top seat is the centre height plus the radius, the bottom seat is the centre height minus it.

Key idea

Distance from the centre to an axis decides everything: equal to $r$ touches, less than $r$ crosses. More than $r$ misses.

Where the circle meets the axes

Every point on the x-axis has $y = 0$ . Every point on the y-axis has $x = 0$ . So to find where a circle cuts an axis, put that coordinate to 0 and solve.

Worked example

1
The circle $(x - 1)^{2} + (y + 2)^{2} = 10$ meets the x-axis at A and B, and the negative y-axis at C. Find the exact area of triangle ABC.
2
$(x - 1)^{2} + 4 = 10$ — Put $y = 0$ .
3
$(x - 1)^{2} = 6$ — Take 4 from both sides.
4
$x = 1 \pm 6$ — Square root. Remember the $\pm$ .
5
$A (1 - 6, 0), B (1 + 6, 0)$ — Two exact points.
6
$1 + (y + 2)^{2} = 10$ — Now put $x = 0$ .
7
$(y + 2)^{2} = 9$ — Take 1 from both sides.
8
$y = 1 or y = - 5$ — C is on the negative y-axis, so $C (0, - 5)$ .
9
$A B = 26$ — Base: the gap between the x-values.
10
$\frac{1}{2} \times 26 \times 5 = 56$ — Height is 5, the distance from C to the x-axis.

Common mistake

Keeping only $+ 6$ . A line through a circle usually cuts it twice.

Rule

“Exact area” means surds, not decimals. On a “show all stages of your working” question, a calculator answer alone scores nothing.

Key idea

Axis = coordinate 0. Solve, keep both roots, then use the points for lengths and areas.

An unknown point on the circle

Sometimes one coordinate of a point on the circle is unknown. Two routes work. Pick whichever looks shorter.

Worked example

1
$Q (- 3, 4)$ and $R (5, - 2)$ are the ends of a diameter. $P (0, p)$ lies on the circle, with $p > 0$ . Find the exact value of $p$ .
2
$Centre (1, 1)$ — Midpoint of QR.
3
$r^{2} = (5 - 1)^{2} + (- 2 - 1)^{2} = 25$ — Centre to R.
4
$(0 - 1)^{2} + (p - 1)^{2} = 25$ — P is on the circle, so its coordinates fit.
5
$(p - 1)^{2} = 24$ — Take 1 from both sides.
6
$p = 1 \pm 26$ — Square root, and $24 = 26$ .
7
$p = 1 + 26$ — We are told $p > 0$ , so reject the negative one.

Rule

Second route: the angle in a semicircle is $9 0^{\circ}$ . So angle QPR is a right angle, and the gradients of PQ and PR multiply to $- 1$ .

Worked example

1
Same question, using gradients.
2
$\frac{p - 4}{0 - ( - 3 )} \times \frac{p - ( - 2 )}{0 - 5} = - 1$ — Gradient PQ times gradient PR.
3
$(p - 4) (p + 2) = 15$ — Multiply both sides by $- 15$ .
4
$p^{2} - 2 p - 23 = 0$ — Expand and collect.
5
$p = 1 + 26$ — Complete the square: $(p - 1)^{2} = 24$ again. Keep $p > 0$ .

Common mistake

Using the length of QR as the radius. QR is the diameter. Find $r^{2}$ from the centre to one end.

Common mistake

Giving both values of $p$ . Use $p > 0$ to keep only one.

Key idea

A point on the circle fits the equation. A point on a diameter's circle also makes a right angle. Either fact gives you an equation to solve.

Which method? Read the question

The wording tells you which tool to use.

You see	You do
Centre and radius “in terms of k”	Flip the signs, square root the right side
A point lies on the circle	Substitute it, solve for the unknown
“Possible range of k”	Write $r^{2}$ in terms of k, then $r^{2} > 0$
“Touches the x-axis” (or y-axis)	Radius = distance from centre to that axis
Meets the axes	Put $y = 0$ or $x = 0$ , keep both roots
Nearest, highest, lowest point	Move one radius from the centre
A diameter and a point on the circle	Substitute into the equation, or gradients multiply to $- 1$

Common mistake

Using rounded decimals. Every answer in this lesson is exact: keep surds to the end.

Key idea

Find the centre and $r^{2}$ first, even with a $k$ in them. Every question here starts from those two.

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Real past-paper questions on circles with unknowns and axes

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