Tangent to a Circle with an Off-Origin Centre

The tangent method doesn’t change for an off-origin circle: it’s still perpendicular to the radius at the point of contact. You just take the radius gradient from the actual centre. (Compare the origin-centred version.)

A circle has equation $(x - 1)^2 + (y - 2)^2 = 25$. Find the equation of the tangent at the point $P(4, 6)$, giving your answer in the form $ax + by + c = 0$. [5]

The working

Step 1 (worth doing), confirm $P$ is on the circle. Centre $C(1, 2)$; $CP = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = 5$, matching the radius. ✓

Step 2, gradient of the radius $CP$ (from the centre $(1, 2)$, not the origin): $m_{CP} = \frac{6 - 2}{4 - 1} = \frac{4}{3} \quad \text{(M1)}$

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

Step 4, tangent through $P(4, 6)$: $y - 6 = -\frac{3}{4}(x - 4) \quad \text{(M1)}$

Clear the fraction and rearrange: $4(y - 6) = -3(x - 4) \;\Rightarrow\; 4y - 24 = -3x + 12 \;\Rightarrow\; 3x + 4y - 36 = 0 \quad \text{(A1, A1)}$

Where the marks are won and lost

• Read the centre as $(1, 2)$ from the equation and use it, not the origin, for the radius gradient. This is the only difference from an origin-centred circle.
• Tangent gradient is the negative reciprocal of $\frac43$, i.e. $-\frac34$.
• The tangent passes through $P$, not the centre; and clear the fraction for the integer form.

Common mistakes

• Using the origin instead of the centre $(1, 2)$ for the radius gradient.
• Using the radius gradient (not its negative reciprocal) for the tangent.
• Writing the line through the centre rather than through $P$.

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