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It might be useful to look at M3 Geometryand area of sectors and M8 Surds

Any point P with coordinates (\(x, y\)) on the circumference of a circle can be joined to the centre (0, 0) by a straight line that forms the hypotenuse of a right angle triangle with sides of length x and y.

This means that, using Pythagoras’ theorem, the equation of a circle with radius r and centre (0, 0) is given by the formula \(x^2 + y^2 = r^2\).

Example

Find the equation of a circle with radius 3 units and centre (0, 0)

The radius, \( r = 3\) and \( r^2 = 9\), so the equation of the circle is \(x^2 + y^2 = 9\)

Example

What is the radius of the circle given by the equation \(x^2 + y^2 = 15\)

The value of \( r^2=15\)so the radius of the circle is \(\sqrt{15} = 3.872983346…\)

This answer can be left as a surd to give an exact answer or, rounded to 1 decimal place, the length is 3.9 units.

A tangent to a circle at point P with coordinates \((x, y)\) 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 values of \(x\) and \(y\) 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).

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 \(\frac{\text{change in y}}{\text{change in x}} = \frac{-4}{3}\).

  • 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 \(y = \frac{3}{4}x - \frac{25}{4}\)