Science:Math Exam Resources/Courses/MATH105/April 2013/Question 05 (a)/Solution 1

Examine the picture below

The red circle describes the contaminated region. Our blue circle has to be just large enough to completely contain the red circle. Hence, we need to find the distance between any point on the red circle and the centre of the blue circle, (2,2). The radius of the blue circle is then the smallest number that is bigger than all these distances.

The distance between any point (x,y) and (2,2) is given by $\displaystyle {\sqrt {(x-2)^{2}+(y-2)^{2}}}$. Hence, the radius r we are looking for is the maximum value of distances from points on the red circle to the centre of the blue circle:

$\displaystyle r={\sqrt {(x-2)^{2}+(y-2)^{2}}}$

subject to this point $(x,y)$ lying on the red circle

$\displaystyle x^{2}+y^{2}=16$

Maximizing this distance gives the smallest possible radius such that the blue circle still encloses the entire contamination area.