Cambridge Syllabus Notes

Coordinate Geometry of the Circle: Tangents

🔵 What is a tangent? A tangent to a circle is a straight line that touches the circle at exactly one point. Imagine a bicycle wheel and a road that just kisses the rim – that road is the tangent.

🧭 Why care? In exams you’ll often need to write the equation of a tangent or prove that a line is tangent. Knowing the rules makes it quick and easy.

Key Concepts

Circle equation: $(x-h)^2 + (y-k)^2 = r^2$
Radius to point of tangency is perpendicular to the tangent line.
For a point $(x_0,y_0)$ on the circle, slope of radius: $\displaystyle \frac{y_0-k}{x_0-h}$
Slope of tangent: negative reciprocal of radius slope.
Equation of tangent at $(x_0,y_0)$: $$(x_0-h)(x-h) + (y_0-k)(y-k) = r^2$$ (simplifies to a linear equation).

Finding the Tangent at a Point on the Circle

Identify the circle’s centre $(h,k)$ and radius $r$.
Take the point of tangency $(x_0,y_0)$.
Use the formula: $$(x_0-h)(x-h) + (y_0-k)(y-k) = r^2$$
Expand and simplify to get the line’s standard form $Ax + By + C = 0$.

🔍 Example: Circle $(x-3)^2 + (y+2)^2 = 25$; point $P(6,1)$.

Step	Calculation
Centre & radius	$(h,k)=(3,-2)$, $r=5$
Tangent formula	$(6-3)(x-3)+(1+2)(y+2)=25$
Simplify	$3(x-3)+3(y+2)=25 \;\Rightarrow\; 3x+3y-9+6=25 \;\Rightarrow\; x+y=10$

So the tangent line is $x + y = 10$.

Tangents from an External Point

When a point $P$ lies outside the circle, there are two tangents that can be drawn. To find their equations:

Let the circle be $(x-h)^2 + (y-k)^2 = r^2$.
Let $P(x_1,y_1)$ be the external point.
Use the condition that the distance from $P$ to the centre equals the length of the tangent segment: $$(x_1-h)^2 + (y_1-k)^2 = r^2 + t^2$$ where $t$ is the length of the tangent.
Alternatively, use the point–tangent form: $$(x_1-h)(x-h) + (y_1-k)(y-k) = r^2$$ which gives a quadratic in $x$ or $y$; factor to find the two tangent lines.

🧮 Quick trick: For a circle centred at the origin $(0,0)$, the tangent from $(x_1,y_1)$ has equation $x x_1 + y y_1 = r^2$.

Exam Tips Box

?? Tip 1: Always check if the point lies on the circle before using the point–tangent formula. If it doesn’t, you’re dealing with an external point.

?? Tip 2: For circles not centred at the origin, shift coordinates first: let $X = x-h$, $Y = y-k$. This simplifies calculations.

?? Tip 3: When asked to prove that a line is tangent, show that the line intersects the circle at exactly one point (discriminant zero) or that the radius is perpendicular to the line.

?? Tip 4: Practice converting between point–slope, slope–intercept, and standard forms – examiners may ask for any.

Practice Problem

🔵 Circle: $x^2 + y^2 = 16$. Find the equations of the two tangents from point $P(5,0)$.

🧩 Solution outline:

Use point–tangent form: $5x + 0y = 16 \;\Rightarrow\; 5x = 16 \;\Rightarrow\; x = \frac{16}{5}$ (one tangent).
For the second tangent, note that the circle is symmetric about the x‑axis, so the other tangent is $x = \frac{16}{5}$ as well? Wait! Actually, because $P$ lies on the x‑axis, the two tangents are symmetric above and below the axis. Use the quadratic method: $(5-x)^2 + y^2 = 16 + t^2$ → solve for $y$ to find slopes.

📝 Full solution will be in the next lesson.

Solve problems involving tangents to a circle, including finding equations of tangents