A circle C with centre at the point (2, –1) passes through the point A at (4, –5).....

(a) Find an equation for the circle C.

Before tackling a question like this it is always a good idea to draw a rough sketch of all the information that's given to you. 

The general equation of a circle is x2+y2=r2, where r is the radius of the circle. 

By using the coordinates given to you for the centre of the circle and the rules of graph transformations you can come to the equation: (x-2)+ (y+1)2 = r2.

You can then use Pythagoras' Theorem to obtain a value for r2. Sketching a small triangle between the two given points can be helpful, with the hypotenuse labelled 'r'. The other lengths of the triangle can be found by subtracting respective x and y coordinates for the centre from the coordinates of point A. Applying Pythagoras' Theorem should give you a value for r2 of 20. 

(b) Find an equation of the tangent to the circle C at the point A, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

There are a few ways of going about this question but one of the simpliest is to recognise that the tangent to point A is perpendicular to the line between A and the centre of the circle. You can easily find the gradient of the line by the following: (-5 - -1) / (4 - 2)= -4 / 2 = -2. The gradient of the tangent is then equal to -1 divided by that value, so 1/2. 

Using the gradient of the tangent and the coordinates for point A you can input them into the general equation of a line, obtaining: (y + 5) = 1/2(x - 4). This can be expanded and rearranged to get an answer of: x - 2y - 14 =0

Brodie W. A Level Economics tutor, A Level Maths tutor, GCSE Maths tutor

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