Intersections of lines and conics; tangents and normals

Covers AS91573

Where a line meets a conic

To find the points where a line meets a conic, solve their equations simultaneously.
Substitute the line (e.g. $y = mx + c$ ) into the conic equation.
This produces a quadratic in one variable — solve it.
The number of real solutions tells you how the line and conic meet.

A circle with a secant line and a tangent line

Reading the discriminant

The quadratic $Ax^2 + Bx + C = 0$ has discriminant $\Delta = B^2 - 4AC$ .
$\Delta > 0$ — two solutions: the line is a secant (cuts the conic at two points).
$\Delta = 0$ — one repeated solution: the line is a tangent (touches at one point).
$\Delta < 0$ — no real solutions: the line misses the conic.

Testing or finding a tangent

To show a line is a tangent, substitute and check that $\Delta = 0$ .
To find the value that makes a line a tangent, set $\Delta = 0$ and solve for the unknown ( $m$ or $c$ ).
The point of contact comes from the repeated root: $x = -\dfrac{B}{2A}$ .

Tangents and normals to a circle

A tangent to a circle is perpendicular to the radius at the point of contact.
Gradient of radius $=$ (rise over run) from centre to the point.
Gradient of tangent $= -\dfrac{1}{\text{gradient of radius}}$ (negative reciprocal).
The normal is the line through the point and the centre — it has the same gradient as the radius.

Worked ExampleLine meeting a circle

Find the points where the line $y = x + 1$ meets the circle $x^2 + y^2 = 25$ .

Step 1 — Substitute the line into the circle

$x^2 + (x + 1)^2 = 25$

Step 2 — Expand and simplify to a quadratic

$x^2 + x^2 + 2x + 1 = 25$

$2x^2 + 2x - 24 = 0 \;\Rightarrow\; x^2 + x - 12 = 0$

Step 3 — Solve the quadratic

$(x + 4)(x - 3) = 0 \;\Rightarrow\; x = -4 \text{ or } x = 3$

Step 4 — Find the matching $y$ -values

Using $y = x + 1$ : at $x = -4$ , $y = -3$ ; at $x = 3$ , $y = 4$ .

$\boxed{(-4, -3) \text{ and } (3, 4)}$

Worked ExampleFinding a tangent by the discriminant

Find the values of $c$ for which the line $y = 2x + c$ is a tangent to the circle $x^2 + y^2 = 20$ .

Step 1 — Substitute the line into the circle

$x^2 + (2x + c)^2 = 20$

Step 2 — Form the quadratic in $x$

$x^2 + 4x^2 + 4cx + c^2 - 20 = 0$

$5x^2 + 4cx + (c^2 - 20) = 0$

Step 3 — Set the discriminant to zero for a tangent

With $A = 5$ , $B = 4c$ , $C = c^2 - 20$ :

$\Delta = (4c)^2 - 4(5)(c^2 - 20) = 0$

$16c^2 - 20c^2 + 400 = 0 \;\Rightarrow\; -4c^2 + 400 = 0$

Step 4 — Solve for $c$

$c^2 = 100 \;\Rightarrow\; c = \pm 10$

$\boxed{c = 10 \text{ or } c = -10}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Find the points of intersection of the line $y = x$ and the circle $x^2 + y^2 = 8$ .

Merit

Show that the line $y = 3x - 10$ is a tangent to the circle $x^2 + y^2 = 10$ , and find the point of contact.

Excellence

The point $P(3, 4)$ lies on the circle $x^2 + y^2 = 25$ . Find the equation of the tangent to the circle at $P$ , giving your answer in the form $y = mx + c$ , and justify your method.

Intersections of lines and conics; tangents and normals

Covers AS91573

Where a line meets a conic

To find the points where a line meets a conic, solve their equations simultaneously.
Substitute the line (e.g. $y = mx + c$ ) into the conic equation.
This produces a quadratic in one variable — solve it.
The number of real solutions tells you how the line and conic meet.

A circle with a secant line and a tangent line

Reading the discriminant

The quadratic $Ax^2 + Bx + C = 0$ has discriminant $\Delta = B^2 - 4AC$ .
$\Delta > 0$ — two solutions: the line is a secant (cuts the conic at two points).
$\Delta = 0$ — one repeated solution: the line is a tangent (touches at one point).
$\Delta < 0$ — no real solutions: the line misses the conic.

Testing or finding a tangent

To show a line is a tangent, substitute and check that $\Delta = 0$ .
To find the value that makes a line a tangent, set $\Delta = 0$ and solve for the unknown ( $m$ or $c$ ).
The point of contact comes from the repeated root: $x = -\dfrac{B}{2A}$ .

Tangents and normals to a circle

A tangent to a circle is perpendicular to the radius at the point of contact.
Gradient of radius $=$ (rise over run) from centre to the point.
Gradient of tangent $= -\dfrac{1}{\text{gradient of radius}}$ (negative reciprocal).
The normal is the line through the point and the centre — it has the same gradient as the radius.

Worked ExampleLine meeting a circle

Find the points where the line $y = x + 1$ meets the circle $x^2 + y^2 = 25$ .

Step 1 — Substitute the line into the circle

$x^2 + (x + 1)^2 = 25$

Step 2 — Expand and simplify to a quadratic

$x^2 + x^2 + 2x + 1 = 25$

$2x^2 + 2x - 24 = 0 \;\Rightarrow\; x^2 + x - 12 = 0$

Step 3 — Solve the quadratic

$(x + 4)(x - 3) = 0 \;\Rightarrow\; x = -4 \text{ or } x = 3$

Step 4 — Find the matching $y$ -values

Using $y = x + 1$ : at $x = -4$ , $y = -3$ ; at $x = 3$ , $y = 4$ .

$\boxed{(-4, -3) \text{ and } (3, 4)}$

Worked ExampleFinding a tangent by the discriminant

Find the values of $c$ for which the line $y = 2x + c$ is a tangent to the circle $x^2 + y^2 = 20$ .

Step 1 — Substitute the line into the circle

$x^2 + (2x + c)^2 = 20$

Step 2 — Form the quadratic in $x$

$x^2 + 4x^2 + 4cx + c^2 - 20 = 0$

$5x^2 + 4cx + (c^2 - 20) = 0$

Step 3 — Set the discriminant to zero for a tangent

With $A = 5$ , $B = 4c$ , $C = c^2 - 20$ :

$\Delta = (4c)^2 - 4(5)(c^2 - 20) = 0$

$16c^2 - 20c^2 + 400 = 0 \;\Rightarrow\; -4c^2 + 400 = 0$

Step 4 — Solve for $c$

$c^2 = 100 \;\Rightarrow\; c = \pm 10$

$\boxed{c = 10 \text{ or } c = -10}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Find the points of intersection of the line $y = x$ and the circle $x^2 + y^2 = 8$ .

Merit

Show that the line $y = 3x - 10$ is a tangent to the circle $x^2 + y^2 = 10$ , and find the point of contact.

Excellence

The point $P(3, 4)$ lies on the circle $x^2 + y^2 = 25$ . Find the equation of the tangent to the circle at $P$ , giving your answer in the form $y = mx + c$ , and justify your method.

Intersections of lines and conics; tangents and normals

Where a line meets a conic

Reading the discriminant

Testing or finding a tangent

Tangents and normals to a circle

Step 1 — Substitute the line into the circle

Step 2 — Expand and simplify to a quadratic

Step 3 — Solve the quadratic

Step 4 — Find the matching yyy-values

Step 1 — Substitute the line into the circle

Step 2 — Form the quadratic in xxx

Step 3 — Set the discriminant to zero for a tangent

Step 4 — Solve for ccc

Test yourself

Practice by grade

Intersections of lines and conics; tangents and normals

Where a line meets a conic

Reading the discriminant

Testing or finding a tangent

Tangents and normals to a circle

Step 1 — Substitute the line into the circle

Step 2 — Expand and simplify to a quadratic

Step 3 — Solve the quadratic

Step 4 — Find the matching yyy-values

Step 1 — Substitute the line into the circle

Step 2 — Form the quadratic in xxx

Step 3 — Set the discriminant to zero for a tangent

Step 4 — Solve for ccc

Test yourself

Practice by grade

Step 4 — Find the matching $y$ -values

Step 2 — Form the quadratic in $x$

Step 4 — Solve for $c$

Step 4 — Find the matching $y$ -values

Step 2 — Form the quadratic in $x$

Step 4 — Solve for $c$