+0

Coordinates

0
143
1

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?

Nov 13, 2021

#1
+117437
+1

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.

https://www.geogebra.org/classic/ax7snpvg

Nov 13, 2021
edited by Melody  Nov 13, 2021
edited by Melody  Nov 13, 2021