Line-Circle Intersections | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Substitute the line into the circle

Rearrange the line to $y=\cdots$ or $x=\cdots$ . A vertical line $x=k$ is already ready to substitute.
Replace that variable in the circle equation.
Expand and collect one quadratic equation.
On a calculator-permitted paper, use the polynomial solver first for an ordinary quadratic. If it gives convenient integer or simple fractional roots, write the corresponding factorisation as your working. If the roots are awkward and exact working is required, use the quadratic formula.
Substitute each root into the line to find the other coordinate. An intersection answer needs both coordinates.

Two roots give two points

Worked example

$y=3x-20$ meets $(x+1)^2+(y-2)^2=85$ . Find the intersection points

(9709/12 F/M 2022 Q6(a))

Answer

(5,-5)\quad\text{and}\quad(8,4)

Swipe left or right to see the whole diagram →

The two real roots correspond to the two points where the line cuts the circle.

Branches that must stay visible

Two quadratic roots normally give two intersection points. Keep both unless a stated region or diagram label excludes one.
Match each $x$ to its own $y$ . Do not pair one x-coordinate with the other root's y-coordinate.
For $x=k$ , substitute $k$ directly into the circle. Do not try to write a vertical line as $y=mx+c$ .

Carry the points into a second task

Your turn9709/13 May/June 2023 Q5

The line $y=x-9$ meets $(x-1)^2+(y+4)^2=40$ at $A$ and $B$ . Find both points, then find the circle with diameter $AB$ .

Show worked answer

Answer

$(x-1)^2+(x-5)^2=40$ , so $x^2-6x-7=0$ . Factorise: $(x+1)(x-7)=0$ .

The points are $(-1,-10)$ and $(7,-2)$ .

Their midpoint is $(3,-6)$ and the radius squared is 32, so the second circle is $(x-3)^2+(y+6)^2=32$ .

Substitute the line into the circle

Rearrange the line to $y=\cdots$ or $x=\cdots$ . A vertical line $x=k$ is already ready to substitute.
Replace that variable in the circle equation.
Expand and collect one quadratic equation.
On a calculator-permitted paper, use the polynomial solver first for an ordinary quadratic. If it gives convenient integer or simple fractional roots, write the corresponding factorisation as your working. If the roots are awkward and exact working is required, use the quadratic formula.
Substitute each root into the line to find the other coordinate. An intersection answer needs both coordinates.

Two roots give two points

Worked example

$y=3x-20$ meets $(x+1)^2+(y-2)^2=85$ . Find the intersection points

(9709/12 F/M 2022 Q6(a))

Answer

(5,-5)\quad\text{and}\quad(8,4)

Swipe left or right to see the whole diagram →

The two real roots correspond to the two points where the line cuts the circle.

Branches that must stay visible

Two quadratic roots normally give two intersection points. Keep both unless a stated region or diagram label excludes one.
Match each $x$ to its own $y$ . Do not pair one x-coordinate with the other root's y-coordinate.
For $x=k$ , substitute $k$ directly into the circle. Do not try to write a vertical line as $y=mx+c$ .

Carry the points into a second task

Your turn9709/13 May/June 2023 Q5

The line $y=x-9$ meets $(x-1)^2+(y+4)^2=40$ at $A$ and $B$ . Find both points, then find the circle with diameter $AB$ .

Show worked answer

Answer

$(x-1)^2+(x-5)^2=40$ , so $x^2-6x-7=0$ . Factorise: $(x+1)(x-7)=0$ .

The points are $(-1,-10)$ and $(7,-2)$ .

Their midpoint is $(3,-6)$ and the radius squared is 32, so the second circle is $(x-3)^2+(y+6)^2=32$ .