Answer to Question #153166 in Analytic Geometry for Kinza tahir

Question

Pro-Build is a large international building company hired to build a hospital to serve three communities in Northern Ontario. The facility should be equidistant from all three communities to show no favouritism and to best serve the area. The layout of the region is already defined on co-ordinate system and the three community centres are located at the following points: (0,6), (0, - 2) , and (12, 2) . (HINT: The light should be located in the centre of the circle that contains the three entrance points. This point is called the CIRCUMCENTRE and it is defined as the intersection point of all three perpendicular bisectors of a triangle within a circle.) a) You are a consultant for Pro-Build and must find the co-ordinates for the new hospital. b) Each unit on the grid is equal to 10 kilometres. Find the actual distance from each of the three communities to this chosen hospital site to verify your solution.

Answer 1

a) Let $C(x_c, y_c)$ be the circumcentre, $R$ be the radius of the cirvle . Then the equation of the circle is

$$(x-x_c)^2+(y-y_c)^2=R^2$$

Point $(0,6)$

$$(0-x_c)^2+(6-y_c)^2=R^2$$

Point $(0,-2)$

$$(0-x_c)^2+(-2-y_c)^2=R^2$$

Point $(12,2)$

$$(12-x_c)^2+(2-y_c)^2=R^2$$

$$\begin{alignedat}{2} x_c^2+ 36-12y_c+y_c^2 = x_c^2+4+4y_c+y_c^2 \\ 144-24x_c+x_c^2+4-4y_c+y_c^2 = x_c^2+4+4y_c+y_c^2 \end{alignedat}$$

$$\begin{alignedat}{2} 16y_c= 32 \\ 24x_c = -8y_c+144 \end{alignedat}$$

$$\begin{alignedat}{2} x_c= \dfrac{16}{3} \\ y_c = 2 \end{alignedat}$$

Point $C\big(\dfrac{16}{3}, 2\big)$

b)

$$(0-\dfrac{16}{3})^2+(6-2)^2=R^2$$

$$R=\dfrac{20}{3}$$

$$\dfrac{16}{3}\times10\ km=\dfrac{160}{3}\ km=53\dfrac{1}{3}\ km$$

The actual distance from each of the three communities is $53\dfrac{1}{3}\ km.$