Length of a Tangent from an External Point

The length of a tangent from an external point uses one fact: the tangent is perpendicular to the radius at the point of contact. That makes a right-angled triangle, radius, tangent, and the centre-to-point line as hypotenuse, so the tangent length falls straight out of Pythagoras.

A circle has equation $x^2 + y^2 = 25$. Find the length of the tangent to the circle from the point $P(8, 6)$. [4]

The working

Step 1, read the centre and radius. $x^2 + y^2 = 25$ is centred at $O(0, 0)$ with radius $r = \sqrt{25} = 5$.

Step 2, find the distance $OP$ from the centre to the external point: $OP = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \quad \text{(M1, A1)}$

Step 3, apply Pythagoras in the right-angled triangle (tangent $\perp$ radius, so $OP$ is the hypotenuse): $\text{tangent}^2 = OP^2 - r^2 = 10^2 - 5^2 = 100 - 25 = 75 \quad \text{(M1)}$ $\text{tangent} = \sqrt{75} = 5\sqrt{3} \approx 8.66 \quad \text{(A1)}$

Where the marks are won and lost

• The key is the right angle between tangent and radius, which makes $OP$ the hypotenuse. So it’s $OP^2 - r^2$, not $r^2 - OP^2$ (the external point is outside, so $OP > r$).
• Simplify the surd: $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$. Leaving $\sqrt{75}$ may cost an exact-form mark.
• First confirm $P$ is outside the circle ($OP = 10 > r = 5$), a tangent from an interior point doesn’t exist.

Common mistakes

• Computing $r^2 - OP^2$ and getting a negative (wrong order).
• Forgetting the perpendicularity, and so not knowing which side is the hypotenuse.
• Not simplifying $\sqrt{75}$.

Full method: Tangents & Circle Properties notes. Topic home: Circle Geometry pillar.