Equation of a Tangent (Edexcel GCSE Maths)

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13 November 2024

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First, make sure you are familiar with equations of straight lines and perpendicular lines
A tangent just touches a circle (but does not cross it)
The tangent at point P is perpendicular to the radius OP
- remember, the gradients of perpendicular lines multiply to -1
  - they are negative reciprocals
So if P is a point $(a, b)$ on the circumference, then the gradient of the radius OP is $\frac{b - 0}{a - 0} = \frac{b}{a}$ .
- Therefore the gradient of the tangent to the circle at P is $- \frac{a}{b}$
From here, use $y = m x + c$ to find the equation of the tangent

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 $P (8, 6)$ .

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

$Gradient of O P = \frac{6 - 0}{8 - 0} = \frac{6}{8} = \frac{3}{4}$

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 $y = m x + c$ to find the value of $c$

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

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

the (exam) results speak for themselves:

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