SQA Higher Maths 2010 Exam paper II Q 3
- Author:
- garythomson
- Topic:
- Circle
To find the equation of the smaller circle.
In order to get the equation of a circle we need to know two things; the centre and the radius. In part (b) of the question we are told that the radius of the large circle is three times that of the smaller circle. How does this help us? NB we need to find a simple solution. The clue for me was the wording "relative to a suitable set of co-ordinate axes".
The crucial thing to see is that the triangles ACB (with red sides) and BDE (with green sides) are similar triangles ie they have corresponding sides in ratio. So if we know that AB (the radius of the larger circle and hypotenuse of the red triangle) is three times BE (the radius of small circle and hypotenuse of the green triangle) that means that AC (the red horizontal line) is three times BD (the green horizontal line) or putting it another way BD is one third AC.
However relative to our set of co-ordinates we can easily calculate AC (the horizontal red side) -- its length is 6 units, and therefore the length of BD is 2 units. Similarly the length of CB (the vertical red side) can be easily calculated to be 6 units, and therefore the length of DE (the vertical green side) is 2 units.
In part (a) we calculated the point B (its referred to as point P in the exam paper) to be the point (-1,4). But now we know the "offset" from a point on the circumference to the centre of the small circle -- it is 2 to the left and 2 up ie (-1+2, 4+2) = (, 6). Using Pythagoras' theorem on the green sided triangle the length of the radius is the square root of (2^2 + 2^2) .
Hence the equation which appears at k.