Let P be a point on the graph of equation, (x-94)^2 + (y-4)^2 = 44^2. What is the shortest possible distance between P and the x-axis?
+118609 Let P be a point on the graph of equation, (x-94)^2 + (y-4)^2 = 44^2. What is the shortest possible distance between P and the x-axis?
The centre of the circle is (94,4) and the radius is 44
So the centre is in the first quadrant.
Let P be the point (x,y)
The domain is [50,138] The range is [ ]
\((x-94)^2+(y-4)^2=44^2\\ (y-4)^2=44^2-(x-94)^2\\ If\;\;y\ge4 \;\;then\\ y=\sqrt{44^2-(x-94)^2}+4\\ \text{Distance to x axis is }\\ d=\sqrt{44^2-(x-94)^2}+4\\~\\ If\;\;y<4 \;\;then\\ y=-\sqrt{44^2-(x-94)^2}+4\\ \text{Distance to x axis is }\\ d=|-\sqrt{44^2-(x-94)^2}+4|\\~\\ \)
Here is an interactive graph.
