The circle, and the relative position of a circle and a line – Points, Straight Lines and Shapes in 2D (Senior 5 Mathematics) | Ganzaa

Points, Straight Lines and Shapes in 2D

Lessons

Senior 5 · Mathematics

Points, Straight Lines and Shapes in 2D

4 lessons

Lesson 4 of 4

The circle, and the relative position of a circle and a line

25 min

Every point at the same distance

A radio mast that reaches 5 km in every direction covers a disc, and the edge of its coverage is a circle (illustrative). With coordinates, "at distance 5 from the mast" becomes an equation, and questions such as "does this straight road enter the coverage, and for how long?" become the line–circle problems of this lesson. In S3 you proved the circle theorems with pure geometry; here the same facts, such as "a tangent is perpendicular to the radius", are used with equations.

Definition

Circle with centre $C (a, b)$ and radius $r \gt 0$ : the set of all points $P$ of the plane with $C P = r$ . By the distance formula, its Cartesian equation is

(x - a)^{2} + (y - b)^{2} = r^{2} .

General form: expanding gives $x^{2} + y^{2} + D x + E y + F = 0$ . Conversely, completing the square shows that this equation has centre $\left(-\frac{D}{2}, -\frac{E}{2}\right)$ and

r^2 = \frac{D^2}{4} + \frac{E^2}{4} - F,

so it is a circle only if this number is positive (if it is 0 the equation describes one point; if negative, no point at all).

Example: the circle with centre $(- 1, 4)$ and radius 3 is $(x + 1)^{2} + (y - 4)^{2} = 9$ , or $x^{2} + y^{2} + 2 x - 8 y + 8 = 0$ .

Worked example 1: centre and radius

Find the centre and radius of $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ .

Solution Complete the square in $x$ and in $y$ :

(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 \quad \Rightarrow \quad (x - 3)^2 + (y + 2)^2 = 25.

The centre is $(3, - 2)$ and the radius is 5. (By the formula: $r^{2} = 9 + 4 + 12 = 25$ .)

Worked example 2: from a diameter, and through three points

(a) Find the circle with diameter $A (- 1, 2)$ , $B (5, 10)$ . (b) Find the circle through $A (5, 7)$ , $B (- 2, 6)$ and $C (6, 0)$ .

Solution

The centre is the midpoint $(2, 6)$ , and $r = \frac{AB}{2} = \frac{\sqrt{36 + 64}}{2} = 5$ : $(x - 2)^{2} + (y - 6)^{2} = 25$ .
Put each point in $x^{2} + y^{2} + D x + E y + F = 0$ : $5D + 7E + F = -74, \qquad -2D + 6E + F = -40, \qquad 6D + F = -36.$ Subtracting pairs: $7 D + E = - 34$ and $- D + 7 E = - 38$ . So $D = 7 E + 38$ , $50 E = - 300$ , $E = - 6$ , $D = - 4$ , $F = - 12$ . The circle is $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ : centre $(2, 3)$ , radius $\sqrt{4 + 9 + 12} = 5$ .

A point, a line and a circle

For a point $P (x_{0}, y_{0})$ , compare $C P$ with $r$ : the sign of $(x_{0} - a)^{2} + (y_{0} - b)^{2} - r^{2}$ tells whether $P$ is inside (negative), on (zero) or outside (positive) the circle.

For a line, let $\delta$ be the distance from the centre to the line (Lesson 3):

Distance	Position	Common points	Substituting the line into the circle gives a quadratic with
$\delta \lt r$	secant	2 (a chord of length $2\sqrt{r^2 - \delta^2}$ )	$\Delta \gt 0$
$\delta = r$	tangent	1 (the point of contact)	$\Delta = 0$
$\delta \gt r$	exterior	0	$\Delta \lt 0$

Figure 1. The circle

(x - 3)^{2} + (y + 2)^{2} = 25

and the lines

d_{1} : x - y + 1 = 0

d_{2} : 3 x + 4 y + 24 = 0

and

d_{3} : y = 6

of Worked example 3. The distance

\delta

from the centre decides:

\delta \lt 5

for

d_{1}

\delta = 5

for

d_{2}

\delta = 8 \gt 5

for

d_{3}

Definition

Secant: a line that cuts a circle at two points. Chord: the segment between those two points.

Tangent to a circle: a line that meets the circle at exactly one point, the point of contact; it is perpendicular to the radius drawn to that point.

Example: the line $y = 3$ is tangent to $x^{2} + y^{2} = 9$ at $(0, 3)$ .

Worked example 3: three lines and one circle

For the circle $(x - 3)^{2} + (y + 2)^{2} = 25$ , find the position of $d_{1} : x - y + 1 = 0$ , $d_{2} : 3 x + 4 y + 24 = 0$ and $d_{3} : y = 6$ . Find the chord cut off by $d_{1}$ and the point of contact of $d_{2}$ .

Solution

$d_{1}$ : $\delta = \frac{|3 + 2 + 1|}{\sqrt{2}} = \frac{6}{\sqrt{2}} \approx 4.24 \lt 5$ : secant. Chord $= 2\sqrt{25 - 18} = 2\sqrt{7} \approx 5.29$ . (Substituting $y = x + 1$ : $(x - 3)^{2} + (x + 3)^{2} = 25$ , so $x^{2} = 3.5$ , $x = \pm\sqrt{3.5}$ : two points.)
$d_{2}$ : $\delta = \frac{|9 - 8 + 24|}{5} = 5 = r$ : tangent. The point of contact is the foot of the perpendicular from $C$ : $\lambda = \frac{25}{25} = 1$ , so $T = (3, - 2) - (3, 4) = (0, - 6)$ .
$d_{3}$ : $\delta = |6 - (-2)| = 8 \gt 5$ : no common point.

Important

Tangent at a point $T$ of the circle: the line through $T$ with normal vector $\overrightarrow{CT}$ .

Length of a tangent from an outside point $P$ : in the right-angled triangle $C T P$ , $PT = \sqrt{CP^2 - r^2}$ . For the circle $x^{2} + y^{2} + D x + E y + F = 0$ , this is $\sqrt{x_0^2 + y_0^2 + Dx_0 + Ey_0 + F}$ .

Worked example 4: tangents

For the same circle, (a) find the tangent at $T (6, 2)$ ; (b) find the length of the tangents from $P (10, 5)$ .

Solution

$T$ is on the circle: $9 + 16 = 25$ . ✓ $\overrightarrow{CT} = (3, 4)$ , so the tangent is $3 (x - 6) + 4 (y - 2) = 0$ , that is $3 x + 4 y - 26 = 0$ .
$C P^{2} = 7^{2} + 7^{2} = 98$ , so $PT = \sqrt{98 - 25} = \sqrt{73} \approx 8.54$ (Figure 2).

Figure 2. The tangent at

T (6, 2)

3 x + 4 y - 26 = 0

. A tangent is perpendicular to the radius at the point of contact. From an outside point

P

there are two tangents, of equal length

\sqrt{CP^2 - r^2}

Warning

Not every $x^2 + y^2 + \dots = 0$ is a circle. Check that $\frac{D^2}{4} + \frac{E^2}{4} - F \gt 0$ : $x^{2} + y^{2} - 2 x + 4 y + 10 = 0$ gives $1 + 4 - 10 = - 5$ , so no point satisfies it. Also, the coefficients of $x^{2}$ and $y^{2}$ must be equal, and there must be no $x y$ term; divide by the common coefficient first, as in $2x^2 + 2y^2 - 8x = 0 \to x^2 + y^2 - 4x = 0$ .

Tip

Distance first, algebra second. To decide the position of a line and a circle, the distance $\delta$ from the centre is quicker than solving a quadratic. Solve only when you need the actual points.

bprp math basics: Equation of a circle (standard form and completing the square). Watch how completing the square in x and in y turns the general equation into the form that shows the centre and the radius.

✦

Practice

The last two are harder.

Write, in both forms, the equation of the circle with centre $(- 2, 5)$ and radius 4.
Find the centre and radius of $x^{2} + y^{2} + 8 x - 2 y - 8 = 0$ .
Is $x^{2} + y^{2} - 2 x + 4 y + 10 = 0$ a circle?
Find the circle with diameter $A (2, - 3)$ , $B (- 4, 5)$ .
For the circle $(x - 1)^{2} + (y - 5)^{2} = 16$ , say whether $(4, 4)$ , $(1, 9)$ and $(0, 0)$ are inside, on or outside it.
Find the position of $3 x - 4 y + 5 = 0$ relative to $x^{2} + y^{2} = 4$ , and the length of any chord.
Find the tangent to $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ at the point $(5, 1)$ .
For which values of $k$ does the line $y = x + k$ touch the circle $x^{2} + y^{2} = 8$ ? Find the points of contact.

Show answersHide answers

$(x + 2)^{2} + (y - 5)^{2} = 16$ ; $x^{2} + y^{2} + 4 x - 10 y + 13 = 0$ .
$(x + 4)^{2} + (y - 1)^{2} = 25$ : centre $(- 4, 1)$ , radius 5.
No: $r^2 = 1 + 4 - 10 = -5 \lt 0$ ; no point satisfies it.
Centre $(- 1, 1)$ , $r = \frac{\sqrt{36 + 64}}{2} = 5$ : $(x + 1)^{2} + (y - 1)^{2} = 25$ .
$(4, 4)$ : $9 + 1 = 10 \lt 16$ , inside. $(1, 9)$ : $0 + 16 = 16$ , on. $(0, 0)$ : $1 + 25 = 26 \gt 16$ , outside.
$\delta = \frac{|5|}{5} = 1 \lt 2$ : secant; chord $2\sqrt{4 - 1} = 2\sqrt{3} \approx 3.46$ .
Centre $(2, - 3)$ , $r = 5$ ; $(5, 1)$ is on it since $9 + 16 = 25$ . Normal $(3, 4)$ : $3 x + 4 y - 19 = 0$ .
Tangent when $\frac{|k|}{\sqrt{2}} = \sqrt{8}$ , so $∣ k ∣ = 4$ : $k = 4$ touches at $(- 2, 2)$ , $k = - 4$ at $(2, - 2)$ .

1Every point at the same distance

2A point, a line and a circle

Every point at the same distance

A point, a line and a circle