A chord of a circle is a straight line segment between any two points on the circle.

Define perpendicular bisector .

The perpendicular bisector of a line segment is perpendicular to it and passes through its midpoint .

True or False? Every chord of a circle passes through its centre.

False. Only a diameter does. A chord joins any two points on the circle, so most chords miss the centre.

How can you find the centre of a circle if you know three points on it?

Draw any two chords between the points, then construct the perpendicular bisector of each. The two bisectors meet at the centre .

Define circumcircle .

The circumcircle of a triangle is the unique circle passing through all three of its vertices.

What does the angle in a semicircle property say?

Any angle at the circumference in a semicircle is a right angle . Equivalently, if a triangle is right-angled, its hypotenuse is a diameter of its circumcircle.

True or False? Every triangle has a circumcircle.

True. It is always possible to draw a unique circle through the three vertices of any triangle.

Where is the centre of the circumcircle of a right-angled triangle?

At the midpoint of the hypotenuse , because the hypotenuse is a diameter.

How can you show that a triangle is right-angled?

Two ways, depending on what you are given. If you know the side lengths , show that they satisfy Pythagoras' theorem . If the three vertices lie on a circle, show that one side is a diameter , since the angle in a semicircle is a right angle.

Define tangent (to a circle).

A tangent is a line that touches a circle at a single point without cutting across it.

How is a tangent related to the radius at the point where it touches the circle?

The tangent is perpendicular to that radius.

International A Level (IAL)MathsEdexcelPure 2FlashcardsCoordinate Geometry in the (x, y) PlaneCircles

Circles (Edexcel International A Level (IAL) Maths: Pure 2): Flashcards

Exam code: YMA01

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FrontEquation of a Circle
Complete the equation of a circle with centre $(a, b)$ and radius $r$ :
${(x - a)}^{2} + {(______)}^{2} =______$

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Cards in this collection (28)

Complete the equation of a circle with centre $(a, b)$ and radius $r$ :
${(x - a)}^{2} + {(______)}^{2} =______$
The completed equation is:
${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$
The right-hand side is the radius squared.
A circle has centre $(2, - 5)$ and radius $3$ . What is its equation?
${(x - 2)}^{2} + {(y + 5)}^{2} = 9$ .
The $- 5$ turns into $+ 5$ inside the bracket, because subtracting a negative number adds it.
True or False?
The equation $x^{2} + y^{2} = 16$ describes a circle.
True.
It is the case where the centre is the origin, so $a$ and $b$ are both zero and each bracket collapses to a single letter.
A circle centred anywhere else keeps its brackets.
A circle has centre $(- 5, - 7)$ and passes through the point $(1, 1)$ . What has to be worked out before its equation can be written down?
The radius, which is the distance from the centre to that point.
Here the steps across and up are $6$ and $8$ , so $r = \sqrt{6^{2} + 8^{2}} = 10$ and the equation is ${(x + 5)}^{2} + {(y + 7)}^{2} = 100$ .
Must the radius of a circle be a whole number?
No. Square-rooting the right-hand side of the equation often leaves a surd.
For ${(x - 1)}^{2} + {(y + 6)}^{2} = 20$ the radius is $\sqrt{20}$ , which simplifies to $2 \sqrt{5}$ .
Why does the equation of a circle have two squared brackets added together?
Because it is Pythagoras' theorem in disguise: $(x - a)$ and $(y - b)$ are the steps across and up from the centre to a point on the circle.
Those two steps are the short sides of a right-angled triangle whose hypotenuse is the radius, so their squares add to $r^{2}$ .
What do you do when a circle equation is not in the form ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ ?
Rearrange it into that form, usually by completing the square. That is the form the centre and radius can be read off from.
Completing the square: $x^{2} - 6 x = {(x - 3)}^{2} -______$
$x^{2} - 6 x = {(x - 3)}^{2} - 9$
What is the general form of the equation of a circle?
$x^{2} + y^{2} + 2 f x + 2 g y + c = 0$
From the general form $x^{2} + y^{2} + 2 f x + 2 g y + c = 0$ , what are the centre and radius?
Centre $(- f, - g)$ , radius $\sqrt{f^{2} + g^{2} - c}$
Find the centre and radius of the circle $x^{2} - 6 x + y^{2} + y - 15 = 0$ .
Completing the square gives ${(x - 3)}^{2} + {(y + \frac{1}{2})}^{2} = \frac{97}{4}$
Centre $(3, - \frac{1}{2})$ , radius $\frac{\sqrt{97}}{2}$
Define chord.
A chord of a circle is a straight line segment between any two points on the circle.
What is the midpoint of the line segment joining $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ ?
$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
Define perpendicular bisector.
The perpendicular bisector of a line segment is perpendicular to it and passes through its midpoint.
Complete the formula for the gradient of the perpendicular bisector of the line segment joining $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , by filling in the denominator:
$- (\frac{x_{2} - x_{1}}{______})$
$- (\frac{x_{2} - x_{1}}{y_{2} - y_{1}})$
The $x$ and $y$ differences swap places compared with the gradient of the segment itself.
What does the perpendicular bisector of a chord always pass through?
The centre of the circle.
True or False?
Every chord of a circle passes through its centre.
False.
Only a diameter does. A chord joins any two points on the circle, so most chords miss the centre.
How can you find the centre of a circle if you know three points on it?
Draw any two chords between the points, then construct the perpendicular bisector of each. The two bisectors meet at the centre.
Define circumcircle.
The circumcircle of a triangle is the unique circle passing through all three of its vertices.
What does the angle in a semicircle property say?
Any angle at the circumference in a semicircle is a right angle.
Equivalently, if a triangle is right-angled, its hypotenuse is a diameter of its circumcircle.
True or False?
Every triangle has a circumcircle.
True.
It is always possible to draw a unique circle through the three vertices of any triangle.
Where is the centre of the circumcircle of a right-angled triangle?
At the midpoint of the hypotenuse, because the hypotenuse is a diameter.
For a right-angled triangle, the radius of its circumcircle is $______$ the length of the hypotenuse.
For a right-angled triangle, the radius of its circumcircle is half the length of the hypotenuse.
How can you show that a triangle is right-angled?
Two ways, depending on what you are given.
If you know the side lengths, show that they satisfy Pythagoras' theorem.
If the three vertices lie on a circle, show that one side is a diameter, since the angle in a semicircle is a right angle.
Define tangent (to a circle).
A tangent is a line that touches a circle at a single point without cutting across it.
How is a tangent related to the radius at the point where it touches the circle?
The tangent is perpendicular to that radius.
The gradient of a tangent to a circle is the negative $______$ of the gradient of the radius at the point where the tangent touches.
The gradient of a tangent is the negative reciprocal of the gradient of the radius at that point.
For example, if the radius has gradient $\frac{2}{3}$ , the tangent has gradient $- \frac{3}{2}$ .
What do you need in order to write down the equation of a tangent at a point $P$ on a circle?
The gradient of the tangent, which is perpendicular to the radius at $P$ , and the coordinates of $P$ .

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