Euclid and Apollonius stand on two seperate portions which Apollonius 12 meters due worth and 9 meters due East of Euclid. Meanw
hile, phytagoras is jogging in a path in such away that his distance from Apollonius is also always twice that from Euclid. Show that Phytagoras' path is circular.
This is connected on how to find circle please show your solution. Thank you so much!
This is connected on how to find circle please show your solution. Thank you so much!
1 answer:
Draw a diagram to illustrate the problem as shown in the figure below.
Euclid is placed at the origin at (0,0).
Apollonius is 12 m north and 9 m east of Euclid, so its coordinate is (9,12).
Pythagoras is at the arbitrary position (x,y) so that is is at distance d from Euclid and 2d from Apollonius.
From the distance formula, obtain
d² = x² + y² (1)
(2d)² = (x-9)² + (y-12)²
or
4d² = (x-9)² + (y-12)² (2)
Substitute (1) into (2).
4(x² + y²) = x² - 18x + 81 + y² - 24y + 144
3x² + 3y² + 18x + 24y = 225
Divide by 3.
x² + 6x + y² + 8y = 75
Create perfect squares.
(x+3)² - 9 + (y+4)² - 16 = 75
(x+3)² + (y+4)² = 10²
Answer:
The path of Pythagoras is a circle of radius 10 m, centered at (-3, -4).
Euclid is placed at the origin at (0,0).
Apollonius is 12 m north and 9 m east of Euclid, so its coordinate is (9,12).
Pythagoras is at the arbitrary position (x,y) so that is is at distance d from Euclid and 2d from Apollonius.
From the distance formula, obtain
d² = x² + y² (1)
(2d)² = (x-9)² + (y-12)²
or
4d² = (x-9)² + (y-12)² (2)
Substitute (1) into (2).
4(x² + y²) = x² - 18x + 81 + y² - 24y + 144
3x² + 3y² + 18x + 24y = 225
Divide by 3.
x² + 6x + y² + 8y = 75
Create perfect squares.
(x+3)² - 9 + (y+4)² - 16 = 75
(x+3)² + (y+4)² = 10²
Answer:
The path of Pythagoras is a circle of radius 10 m, centered at (-3, -4).
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