Equations of Tangents to Circles

Video lesson · from 26:371 worked example

One lesson video covers all of Circles, so it opens at 26:37 for this topic — not from the beginning.

We know from National 5 that if we have a circle and a tangent to that circle, the tangent meets the radius at right angles.

If the point of contact between the circle and the tangent is known, then we can calculate the gradient of the radius.

The gradient of the tangent can then be found using $m_{radius} \times m_{tangent} = -1$ .

We then know the gradient of the tangent and we have the point of contact, therefore can find the equation of the tangent using $y - b = m(x - a)$ .

Radius and tangent are perpendicular: so the tangent's gradient is the negative reciprocal of the radius gradient. Using the radius gradient itself is the standard error.
Find the radius gradient first: from the centre to the point of contact — in that order, though the gradient is the same either way.
Use the point of contact in the equation: $y - b = m(x - a)$ needs the point on the circle, not the centre.
"Show that the line is a tangent" is different: substitute and show the discriminant is zero, rather than finding an equation.

Example 1

A(1, 3) lies on the circle with equation $x^2+y^2+6x+2y-22=0$ . Find the equation of the tangent at A.

1. Find the centre of the circle:

2g = 6 \implies g=3, \quad 2f = 2 \implies f=1

Centre C = $(-3, -1)$

2. Calculate the gradient of the radius (from C to A):

m_{radius} = \frac{3 - (-1)}{1 - (-3)} = \frac{4}{4} = 1

3. Calculate the gradient of the tangent:

Since the tangent is perpendicular to the radius, $m_{tangent} = -1 / m_{radius} = -1$ .

4. Find the equation of the tangent passing through A(1, 3):

y - 3 = -1(x - 1)

y - 3 = -x + 1

y = -x + 4

Or alternatively expressed as $x+y-4=0$ .