In how many ways can a straight line meet a circle?

Three: it can cut the circle twice , touch it once, or miss it entirely. A line cutting twice makes a chord , which is a diameter if it passes through the centre, and a line touching once is a tangent .

How do you find where a line meets a circle?

Substitute the line equation into the circle equation, then expand and rearrange into a quadratic. The number of solutions tells you the answer: two means it cuts twice, one repeated root means it is a tangent, and none means it misses.

Define a tangent to a circle.

A tangent is a line that touches the circle at exactly one point without cutting across it. That single point is called the point of contact.

How is a tangent related to the radius drawn to its point of contact?

They are perpendicular to each other. This is what makes tangent questions solvable: it converts a fact about touching into a fact about gradients.

True or False? The equation of the tangent is found by using the coordinates of the centre of the circle.

False. The centre is used only to work out the gradient of the radius. The tangent itself passes through the point of contact , so those are the coordinates that go into the equation of the line.

In how many ways can two circles meet?

Three: they can cross twice , touch at exactly one point, or not meet at all. Circles that touch are sometimes described as tangent to each other, since they share a common tangent line at that point.

Define the common chord of two circles.

The common chord is the straight line joining the two points where the circles cross. It is a chord of both circles at once, which is why it only exists when the circles meet twice.

What are the two ways of finding the equation of a common chord?

If you have the circle equations , set them equal to each other: the squared terms cancel and what remains is the chord itself, so no coordinates are needed. If you already have the two intersection points , find the gradient between them and use the point-gradient form.

IGCSEFurther MathsCambridge (CIE)Additional MathsFlashcardsCoordinate GeometryCoordinate Geometry of the Circle

Coordinate Geometry of the Circle (Cambridge (CIE) IGCSE Additional Maths): Flashcards

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

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Cards in this collection (24)

Complete the equation of a circle with centre $(a, b)$ and radius $r$ :
$(x -______)^{2} + (y -______)^{2} =______$
The completed equation is:
$(x - a)^{2} + (y - b)^{2} = r^{2}$
The right-hand side is the radius squared, not the radius itself.
True or False?
The circle $(x + 5)^{2} + (y - 1)^{2} = 36$ has centre $(5, - 1)$ .
False.
The centre is $(- 5, 1)$ , found by asking what makes each bracket zero.
The numbers inside the brackets always have the opposite signs to the coordinates of the centre.
Write the equation of the circle with centre $(- 1, 4)$ and radius 3.
Substitute into $(x - a)^{2} + (y - b)^{2} = r^{2}$ , remembering the signs flip:
$(x + 1)^{2} + (y - 4)^{2} = 9$
The radius 3 becomes 9 on the right, because the equation uses $r^{2}$ .
What is the general form of the equation of a circle?
It is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .
In this form the centre and radius are hidden, so it usually has to be rearranged before it is useful.
How do you get from the general form to the centre and radius?
Complete the square separately in $x$ and in $y$ , then tidy the constants onto the right-hand side.
That puts the equation back into $(x - a)^{2} + (y - b)^{2} = r^{2}$ , where the centre and radius can be read off.
Find the centre and radius of the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ .
Completing the square in each variable gives:
$(x - 3)^{2} + (y + 2)^{2} = 25$
So the centre is $(3, - 2)$ and the radius is $5$ , taking the positive square root of 25.
In how many ways can a straight line meet a circle?
Three: it can cut the circle twice, touch it once, or miss it entirely.
A line cutting twice makes a chord, which is a diameter if it passes through the centre, and a line touching once is a tangent.
How do you find where a line meets a circle?
Substitute the line equation into the circle equation, then expand and rearrange into a quadratic.
The number of solutions tells you the answer: two means it cuts twice, one repeated root means it is a tangent, and none means it misses.
Show that $y = 4 x + 4$ is a tangent to $(x - 4)^{2} + (y - 3)^{2} = 17$ .
Substituting and simplifying collapses the quadratic to:
$17 x^{2} = 0$
That has the single repeated solution $x = 0$ , so there is exactly one point of contact, at $(0, 4)$ .
Why is the full substitution method unnecessary for a line such as $y = 7$ ?
Because a horizontal line already tells you the value of $y$ , so you can put 7 straight into the circle equation.
That leaves an equation in $x$ alone, and the same shortcut works for a vertical line $x = k$ .
Define a tangent to a circle.
A tangent is a line that touches the circle at exactly one point without cutting across it.
That single point is called the point of contact.
How is a tangent related to the radius drawn to its point of contact?
They are perpendicular to each other.
This is what makes tangent questions solvable: it converts a fact about touching into a fact about gradients.
How do you find the equation of the tangent at a point $P$ on a circle with centre $O$ ?
Find the gradient of the radius $O P$ , then take its negative reciprocal to get the gradient of the tangent.
Put that gradient and the coordinates of $P$ into $y - y_{2} = m (x - x_{2})$ .
The radius from $(x_{1}, y_{1})$ to $(x_{2}, y_{2})$ has gradient $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ . Complete the tangent gradient:
$m_{t} =______$
The completed formula is:
$m_{t} = - \frac{x_{2} - x_{1}}{y_{2} - y_{1}}$
This is the negative reciprocal written out in full, so it gives the tangent gradient in one step without finding the radius gradient first.
A circle has centre $(1, 2)$ and passes through $P (4, 6)$ . What is the gradient of the tangent at $P$ ?
The radius has gradient $\frac{6 - 2}{4 - 1} = \frac{4}{3}$ .
The tangent is perpendicular to it, so its gradient is the negative reciprocal, $- \frac{3}{4}$ .
True or False?
The equation of the tangent is found by using the coordinates of the centre of the circle.
False.
The centre is used only to work out the gradient of the radius.
The tangent itself passes through the point of contact, so those are the coordinates that go into the equation of the line.
In how many ways can two circles meet?
Three: they can cross twice, touch at exactly one point, or not meet at all.
Circles that touch are sometimes described as tangent to each other, since they share a common tangent line at that point.
What do you compare to decide whether two circles meet?
The distance $d$ between the two centres, against the two radii.
Find $d$ using Pythagoras on the centres, then compare it with the sum and the difference of the radii.
For circles with radii $r_{1}$ and $r_{2}$ , where $r_{2} \geq r_{1}$ , complete the condition for crossing twice:
$r_{2} - r_{1}______d______r_{1} + r_{2}$
The completed condition is:
$r_{2} - r_{1} < d < r_{1} + r_{2}$
The centres must be far enough apart that one circle is not swallowed by the other, but close enough that they still reach each other.
True or False?
Two circles touch at exactly one point only when the distance between their centres equals the sum of their radii.
False.
That is touching on the outside, where the circles sit side by side.
They also touch when $d = r_{2} - r_{1}$ , which is one circle sitting inside the other and meeting it at a single point.
How do you find the coordinates where two circles intersect?
Rearrange both equations so each has zero on one side, then set them equal to each other.
The $x^{2}$ and $y^{2}$ terms cancel, leaving a linear equation that the intersections must lie on.
Substitute that line back into either circle equation to find the coordinates.
Two circles meet where $x = 1$ , and one of them is $x^{2} + y^{2} = 2$ . Find the points of intersection.
Substitute $x = 1$ into the circle equation:
$1 + y^{2} = 2$
So $y^{2} = 1$ , giving $y = 1$ and $y = - 1$ , and the intersections are $(1, 1)$ and $(1, - 1)$ .
Define the common chord of two circles.
The common chord is the straight line joining the two points where the circles cross.
It is a chord of both circles at once, which is why it only exists when the circles meet twice.
What are the two ways of finding the equation of a common chord?
If you have the circle equations, set them equal to each other: the squared terms cancel and what remains is the chord itself, so no coordinates are needed.
If you already have the two intersection points, find the gradient between them and use the point-gradient form.

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