Mathematics 9709 · Pure Mathematics 1

Problems involving intersections of lines and circles

To find where a line meets a circle, put the line into the circle and solve the one quadratic you get.
The discriminant of that quadratic tells you the shape before you solve: cut, touch, or miss. This is how “show this is a tangent” questions are marked.

Put the line in. Do not guess. Put the line into the circle to get one quadratic in $x$ (or $y$ ). Its roots are the crossing points. Find the other coordinate from the line.

The discriminant tells you the shape: $b^2-4ac>0$ cuts (two points), $b^2-4ac=0$ is a tangent (one repeated root), $b^2-4ac<0$ misses.

Substitute, solve, and read the discriminant

Because the line gives $y$ in terms of $x$ , replacing $y$ in the circle leaves only $x$ , so you always end up with a single quadratic in one variable.

Rearrange the line to $y=\dots$ (or $x=\dots$ ) and put it into the circle to get one quadratic; its solutions are the crossing points, but the discriminant tells you the shape first:

$b^2-4ac>0$ → the line cuts the circle at two points.
$b^2-4ac=0$ → the line is a tangent (it touches once).
$b^2-4ac<0$ → the line misses the circle.

Two crossings, one touch (tangent), or a complete miss: the discriminant tells you which.

Two facts make most tangent questions short:

A tangent is perpendicular to the radius at the point where they touch.
The distance from the centre to a tangent equals the radius.

Worked examples

Worked example

Real question: $y = 3x - 20$ meets $(x + 1)^2 + (y - 2)^2 = 85$ at $A$ and $B$ . Find $A$ and $B$

9709/12 Feb/March 2022 Q6(a). Put the line in, then do the algebra with care, because most marks are in the expand-and-collect step.

Substitute the line straight into the circle: $(x+1)^2+(3x-20-2)^2=85$ .
Expand and collect: $10x^2-130x+400=0\;\Rightarrow\;x^2-13x+40=0$ .
Factorise: $(x-5)(x-8)=0\;\Rightarrow\;x=5,\ 8$ .
Get each $y$ from the line $y=3x-20$ : $x=5\Rightarrow y=-5$ , $x=8\Rightarrow y=4$ .

Answer

A=(5,-5),\quad B=(8,4)

The line y = 3x − 20 cuts the circle centre (−1, 2), radius √85 at A(5, −5) and B(8, 4).

Worked example

Show that $y = x - 13$ is a tangent to $x^2 + y^2 - 8x + 6y + 7 = 0$

Cambridge Pure Mathematics 1, Worked example 3.15. “Tangent” means the quadratic has one repeated root (the same root twice).

Substitute $y=x-13$ and simplify to one quadratic: $x^2+(x-13)^2-8x+6(x-13)+7=0$ .
Collect: $2x^2-28x+98=0\;\Rightarrow\;x^2-14x+49=0$ .
Factorise — a perfect square is a single repeated root: $(x-7)^2=0\;\Rightarrow\;x=7\ \text{(repeated)}$ .

Answer

(x-7)^2=0\ \Rightarrow\ \text{the line is a tangent}

From two crossings to the chord and its circle

Why the perpendicular bisector of a chord goes through the centre

A common next part: after the two crossings $A$ and $B$ , the question asks for the circle with $AB$ as a diameter, or for the perpendicular bisector of $AB$ . Now $AB$ is a chord, and the perpendicular bisector of any chord goes through the centre. This is a useful check: the midpoint of $AB$ should line up with the centre.

Where the marks go

Common mistake

A tangent gives one repeated root, so

b^2-4ac=0

, not

>0

. Most lost marks here come from the expand-and-collect step, not from the idea. (9709/12 Mar 2022 Q6)

Common mistake

Give both coordinates of each point. After you have

x

, find

y

from the line. If a diagram marks one point, pick the matching root. (9709/13 Jun 2023 Q5)

Now you try

A circle has equation $(x-1)^2+(y+4)^2=40$ . The line $y=x-9$ meets the circle at $A$ and $B$ . Find the coordinates of $A$ and $B$ , then the equation of the circle with diameter $AB$ . (9709/13 Jun 2023 Q5)

Your turn— tap to reveal the worked answer (9709/13 Jun 2023 Q5)

Put $y=x-9$ into the circle:

$(x-1)^2+(x-5)^2=40\;\Rightarrow\;2x^2-12x-14=0$

$x^2-6x-7=0\;\Rightarrow\;(x-7)(x+1)=0$ , so $x=7$ or $-1$ .

$A=(-1,-10),\ B=(7,-2)$ . Centre = midpoint $(3,-6)$ ; $r^2=\tfrac14 AB^2=32$ .

Answer

A=(-1,-10),\ B=(7,-2);\ (x-3)^2+(y+6)^2=32

PreviousThe equation of a circle NextRadians and converting

Problems involving intersections of lines and circles

To find where a line meets a circle, put the line into the circle and solve the one quadratic you get.
The discriminant of that quadratic tells you the shape before you solve: cut, touch, or miss. This is how “show this is a tangent” questions are marked.

Put the line in. Do not guess. Put the line into the circle to get one quadratic in $x$ (or $y$ ). Its roots are the crossing points. Find the other coordinate from the line.

The discriminant tells you the shape: $b^2-4ac>0$ cuts (two points), $b^2-4ac=0$ is a tangent (one repeated root), $b^2-4ac<0$ misses.

Substitute, solve, and read the discriminant

Because the line gives $y$ in terms of $x$ , replacing $y$ in the circle leaves only $x$ , so you always end up with a single quadratic in one variable.

Rearrange the line to $y=\dots$ (or $x=\dots$ ) and put it into the circle to get one quadratic; its solutions are the crossing points, but the discriminant tells you the shape first:

$b^2-4ac>0$ → the line cuts the circle at two points.
$b^2-4ac=0$ → the line is a tangent (it touches once).
$b^2-4ac<0$ → the line misses the circle.

Two crossings, one touch (tangent), or a complete miss: the discriminant tells you which.

Two facts make most tangent questions short:

A tangent is perpendicular to the radius at the point where they touch.
The distance from the centre to a tangent equals the radius.

Worked examples

Worked example

Real question: $y = 3x - 20$ meets $(x + 1)^2 + (y - 2)^2 = 85$ at $A$ and $B$ . Find $A$ and $B$

9709/12 Feb/March 2022 Q6(a). Put the line in, then do the algebra with care, because most marks are in the expand-and-collect step.

Substitute the line straight into the circle: $(x+1)^2+(3x-20-2)^2=85$ .
Expand and collect: $10x^2-130x+400=0\;\Rightarrow\;x^2-13x+40=0$ .
Factorise: $(x-5)(x-8)=0\;\Rightarrow\;x=5,\ 8$ .
Get each $y$ from the line $y=3x-20$ : $x=5\Rightarrow y=-5$ , $x=8\Rightarrow y=4$ .

Answer

A=(5,-5),\quad B=(8,4)

The line y = 3x − 20 cuts the circle centre (−1, 2), radius √85 at A(5, −5) and B(8, 4).

Worked example

Show that $y = x - 13$ is a tangent to $x^2 + y^2 - 8x + 6y + 7 = 0$

Cambridge Pure Mathematics 1, Worked example 3.15. “Tangent” means the quadratic has one repeated root (the same root twice).

Substitute $y=x-13$ and simplify to one quadratic: $x^2+(x-13)^2-8x+6(x-13)+7=0$ .
Collect: $2x^2-28x+98=0\;\Rightarrow\;x^2-14x+49=0$ .
Factorise — a perfect square is a single repeated root: $(x-7)^2=0\;\Rightarrow\;x=7\ \text{(repeated)}$ .

Answer

(x-7)^2=0\ \Rightarrow\ \text{the line is a tangent}

From two crossings to the chord and its circle

Why the perpendicular bisector of a chord goes through the centre

Where the marks go

Common mistake

A tangent gives one repeated root, so

b^2-4ac=0

, not

>0

. Most lost marks here come from the expand-and-collect step, not from the idea. (9709/12 Mar 2022 Q6)

Common mistake

Give both coordinates of each point. After you have

x

, find

y

from the line. If a diagram marks one point, pick the matching root. (9709/13 Jun 2023 Q5)

Now you try

Your turn— tap to reveal the worked answer (9709/13 Jun 2023 Q5)

Put $y=x-9$ into the circle:

$(x-1)^2+(x-5)^2=40\;\Rightarrow\;2x^2-12x-14=0$

$x^2-6x-7=0\;\Rightarrow\;(x-7)(x+1)=0$ , so $x=7$ or $-1$ .

$A=(-1,-10),\ B=(7,-2)$ . Centre = midpoint $(3,-6)$ ; $r^2=\tfrac14 AB^2=32$ .

Answer

A=(-1,-10),\ B=(7,-2);\ (x-3)^2+(y+6)^2=32