The equation of the tangent to a circle - Higher

A tangent to a circle at point P is a straight line that touches the circle at P. The tangent is perpendicular to the radius which joins the centre of the circle to the point P.

As the tangent is a straight line, the equation of the tangent will be of the form y = mx + c. We can use perpendicular gradients to find the value of m, then use the coordinates of P to find the value of c in the equation.

Example

Find the equation of the tangent to the circle x^2 + y^2 = 25 at the point (3, -4).

Diagram showing the equation of the tangent to the circle x^2 + y^2 = 25 at the point (3, -4).

The tangent will have an equation in the form y = mx + c so to find the equation you need to find the values of m and c. First, find m, the gradient of the tangent.

On a diagram, draw the circle and the tangent at the point P (3, -4) and draw the radius from the centre (0, 0) to the point P. The gradient of the radius is given by \frac{change~in~y}{change~in~x} = - \frac{4}{3}

The radius that joins the centre of the circle (0, 0) to the point P is at right angles to the tangent, so the gradient of the tangent is the negative reciprocal of the gradient of the radius. This means that the gradient of the tangent, m = \frac{3}{4}.

Next, find the value of c. For the point P, the value of x = 3 and the value of y = -4. Substituting these values, as well as the value of m = \frac{3}{4}, in to the formula y = mx + c gives:

y = mx + c

-4 = \frac{3}{4} \times 3 + c

-4 = \frac{9}{4} + c

-4 - \frac{9}{4} = c

c = -\frac{25}{4}

So the equation of the tangent to the circle x^2 + y^2 = 25 at the point (3, -4) is y = \frac{3}{4}x - \frac{25}{4}

Question

Find where the tangent at (2, 4) to the circle, x^2 + y^2 = 20, crosses the x-axis.

Diagram showing the equation of the tangent to the circle x^2 + y^2 = 25 at the point (3, -4).

Sketch a diagram to show the circle and the tangent at the point (2, 4) labelling this P. Draw the radius from the centre of the circle to P. The tangent will have an equation in the form y = mx + c.

First, find m, the gradient of the tangent.

The gradient of the radius is given by \frac{change~in~y}{change~in~x} =\frac{4}{2}= 2.

The gradient of the tangent is the negative reciprocal of the gradient of the radius. This means that the gradient of the tangent, m = - \frac{1}{2}.

Next, find the value of c. For the point P, the value of x = 2 and the value of y = 4. Substituting these values, as well as the value of m = - \frac{1}{2}, in to the formula y = mx + c gives:

y = mx + c

4 = -\frac{1}{2} \times 2 + c

4 = -1 + c

c = 5

So the equation of the tangent y = -\frac{1}{2} x + 5.

Finally, the point where the tangent crosses the x-axis will have a y-coordinate of 0. Substituting this value into the equation for the tangent gives:

0 = -\frac{1}{2}x + 5

-5 = -\frac{1}{2}x

x = 10

So the tangent crosses the x-axis at (10, 0).