This means that, using Pythagoras’ theorem, the equation of a circle with radius \(r\) and centre (0, 0) is given by the formula \(x^2+ y^2= r^2\). Find the equation of a circle with radius 3 ...

To find the equation of a circle when you know the radius and centre, use the formula \({(x - a)^2} + {(y - b)^2} = {r^2}\), where \((a,b)\) represents the centre of the circle, and \(r\) is the ...

This means that, using Pythagoras’ theorem, the equation of a circle with radius r and centre (0, 0) is given by the formula \(x^2 + y^2 = r^2\). Find the equation of a circle with radius 3 ...

You need both a point and the gradient to find its equation. You are usually given the point - it's where the tangent meets the circle. To find the gradient use the fact that the tangent is ...

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 \(y = mx + c\).

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 ...

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 \(y = mx + c\).