Tangent to a Circle at a Given Point

Every “tangent to a circle at a point” question uses one property: the tangent is perpendicular to the radius at the point of contact. Find the radius gradient, flip it to the negative reciprocal, and write the line. It reduces to straight-line geometry.

A circle has centre $C(2, 3)$ and passes through the point $P(5, 7)$. Find the equation of the tangent to the circle at $P$, giving your answer in the form $ax + by + c = 0$. [5]

The working

Step 1 (worth doing), confirm $P$ is on the circle. Radius $CP = \sqrt{(5-2)^2 + (7-3)^2} = \sqrt{9 + 16} = 5$, so $P$ genuinely lies on the circle of radius $5$.

Step 2, gradient of the radius $CP$: $m_{CP} = \frac{7 - 3}{5 - 2} = \frac{4}{3} \quad \text{(M1)}$

Step 3, tangent gradient = negative reciprocal (tangent $\perp$ radius): $m_{\text{tangent}} = -\frac{3}{4} \quad \text{(M1)}$

Step 4, equation of the tangent through $P(5, 7)$: $y - 7 = -\frac{3}{4}(x - 5) \quad \text{(M1)}$

Clear the fraction and rearrange into the requested form: $4(y - 7) = -3(x - 5) \;\Rightarrow\; 4y - 28 = -3x + 15 \;\Rightarrow\; 3x + 4y - 43 = 0 \quad \text{(A1, A1)}$

Where the marks are won and lost

• The perpendicularity is the whole method. The tangent gradient is $-\frac34$ (negative reciprocal of $\frac43$), not $\frac43$ or $-\frac43$.
• Use the point $P$, not the centre $C$, when writing the tangent line, the tangent passes through $P$.
• The demanded form $ax + by + c = 0$ with integer coefficients means clearing the $\frac34$. Leaving $y = -\frac34 x + \dots$ can cost the final mark.

Common mistakes

• Using the radius gradient for the tangent (forgetting the perpendicular step).
• Writing the line through $C(2,3)$ instead of $P(5,7)$.
• Sign or fraction slips when rearranging into $ax + by + c = 0$.

Full method: Tangents & Circle Properties notes. See also Parallel & Perpendicular Lines. Topic home: Circle Geometry pillar.