Equation of a Tangent (Edexcel GCSE Maths: Higher): Revision Note

• Solving simultaneous equations of circle and tangent only gives one solution

  • so if you are asked to show a line is a tangent, solve the simultaneous equations and show there is only one solution

• Always draw a diagram to help!

The graph shows the circle with the equation $x^2+y^2=100$.

Find an equation of the tangent to the circle at the point $P\left(8,6\right)$.

Answer:

Find the gradient of the radius by finding the gradient of the line segment from the origin to the point $P$

$\mathrm{Gradient}\mathrm{of}OP=\frac{6-0}{8-0}=\frac{6}{8}=\frac{3}{4}$

Find the gradient of the tangent by taking the negative reciprocal of the gradient of the radius

$m=-\frac{4}{3}$

Substitute $m=-\frac{4}{3}$, $x=8$ and $y=6$ into $y=mx+c$ to find the value of $c$

$$\begin{gathered} 6=-\frac{4}{3}\times 8+c \\ 6=-\frac{32}{3}+c \\ c=6+\frac{32}{3} \\ c=\frac{50}{3} \end{gathered}$$

The equation of the tangent is $y=-\frac{4}{3}x+\frac{50}{3}$