Question

A circular platform is to be built in a playground. The center of the structure is required to be equidistant from three support columns located at D(−2,−4), E(1,5), and F(2,0). What are the coordinates for the location of the center of the platform?

Answer 1

Answer:

The coordinates for the location of the center of the platform are (-3.5,1.5)

Step-by-step explanation:

You have 3 points:

D(−2,−4)

E(1,5)

F(2,0)

And you have to find a equidistant point (c) ($x_{c}$,$y_{c}$) from the three given.

Then, you know that:

$D_{cD}=D_{cE}$

And:

$D_{cE}=D_{cF}$

Where:

$D_{cD}$=Distance between point c to D

$D_{cE}$=Distance between point c to E

$D_{cF}$=Distance between point c to D

The equation to calculate distance between two points (A to B) is:

$D_{AB}=\sqrt{(x_{B}-x_{A})^2+(y_{B}-y_{A})^2)}$

$D_{AB}=\sqrt{(x_{B}^2)-(2*x_{B}*x_{A})+(x_{A}^2)+(y_{B}^2)-(2*y_{B}*x_{A})+(y_{A}^2)}$

Then you have to calculate:

*$D_{cD}=D_{cE}$

$D_{cD}=\sqrt{(x_{D}-x_{c})^2+(y_{D}-y_{c})^2}$

$D_{cD}=\sqrt{(x_{D}^2)-(2*x_{D}*x_{c})+(x_{c}^2)+(y_{D}^2)-(2*y_{D} y_{c})+(y_{c}^2)}$

$D_{cD}=\sqrt{(-2^2-(2(-2)*x_{c})+x_{c}^2)+(-4^2-(2(-4) y_{c})+y_{c}^2)}$

$D_{cD}=\sqrt{(4+4x_{c}+x_{c}^2 )+(16+8y_{c}+y_{c}^2)}$

$D_{cE}=\sqrt{(x_{E}-x_{c})^2+(y_{E}-y_{c})^2}$

$D_{cE}=\sqrt{(x_{E}^2)-(2*x_{E}*x_{c})+(x_{c}^2)+(y_{E}^2)-(2y_{E}*y_{c})+(y_{c}^2)}$

$D_{cE}=\sqrt{(1^2-2(1)*x_{c}+x_{c}^2)+(5^2-2(5)+y_{c}+y_{c}^2)}$

$D_{cE}=\sqrt{(1-2x_{c}+x_{c}^2)+(25-10y_{c}+y_{c}^2)}$

$D_{cD}=D_{cE}$

$\sqrt{((4+4x_{c}+x_{c}^2)+(16+8y_{c}+y_{c}^2))}=\sqrt{(1-2x_{c}+x_{c}^2)+(25-10y_{c}+y_{c}^2)}$

$(4+4x_{c}+x_{c}^2)+(16+8y_{c}+y_{c}^2)= (1-2x_{c}+x_{c}^2)+(25-10y_{c}+y_{c}^2)$

$x_{c}^2+y_{c}^2+4x_{c}+8y_{c}+20=x_{c}^2+y_{c}^2-2x_{c}-10y_{c}+26$

$4x_{c}+2x_{c}+8y_{c}+10y_{c}=6$

$6x_{c}+18y_{c}=6$

You get equation number 1.

*$D_{cE}=D_{cF}$

$D_{cE}=\sqrt{(x_{E}-x_{c})^2+(y_{E}-y_{c})^2}$

$D_{cE}=\sqrt{(x_{E}^2-(2+x_{E}*x_{c})+x_{c}^2)+(y_{E}^2-(2y_{E} *y_{c})+y_{c}^2)}$

$D_{cE}=\sqrt{((1^2-2(1)+x_{c}+x_{c}^2)+(5^2-2(5)y_{c}+y_{c}^2)}$

$D_{cE}=\sqrt{(1-2x_{c}+x_{c}^2 )+(25-10y_{c}+y_{c}^2)}$

$D_{cF}=\sqrt{(x_{F}-x_{c})^2+(y_{F}-y_{c})^2}$

$D_{cF}=\sqrt{(x_{F}^2-(2*x_{F}*x_{c})+x_{c}^2)+(y_{F}^2-(2*y_{F}* y_{c})+y_{c}^2)}$

$D_{cF}=\sqrt{(2^2-(2(2)x_{c})+x_{c}^2)+(0^2-(2(0)y_{c}+y_{c}^2)}$

$D_{cF}=\sqrt{(4-4x_{c}+x_{c^2})+(0-0+y_{c}^2)}$

$D_{cE}=D_{cF}$

$\sqrt{(1-2x_{c}+x_{c}^2 )+(25-10y_{c}+y_{c}^2)}=\sqrt{(4-4x_{c}+x_{c}^2 )+(0-0+y_{c}^2)}$

$(1-2x_{c}+x_{c}^2)+(25-10y_{c}+y_{c}^2 )=(4-4x_{c}+x_{c}^2)+(0-0+y_{c}^2)$

$x_{c}^2+y_{c}^2-2x_{c}-10y_{c}+26=x_{c}^2+y_{c}^2-4x_{c}+4$

$-2x_{c}+4x_{c}-10y_{c}=-22$

$2x_{c}-10y_{c}=-22$

You get equation number 2.

Now you have to solve this two equations:

$6x_{c}+18y_{c}=6$ (1)

$2x_{c}-10y_{c}=-22$ (2)

From (2)  

$-10y_{c}=-22-2x_{c}$

$y_{c}=(-22-2x_{c})/(-10)$

$y_{c}=2.2+0.2x_{c}$

Replacing $y_{c}$ in (1)

$6x_{c}+18(2.2+0.2x_{c})=6$

$6x_{c}+39.6+3.6x_{c}=6$

$9.6x_{c}=6-39.6$

$x_{c}=6-39.6$

$x_{c}=-3.5$

Replacing $x_{c}=-3.5$ in

$y_{c}=2.2+0.2x_{c}$

$y_{c}=2.2+0.2(-3.5)$

$y_{c}=2.2+0.2(-3.5)$

$y_{c}=2.2-0.7$

$y_{c}=1.5$

Then the coordinates for the location of the center of the platform are (-3.5,1.5)

Answer 2

Answer:

The coordinates for the location of the center of the platform are (-1 , 2)

Step-by-step explanation:

* Lets revise the equation of the circle

- The equation of the circle of center (h , k) and radius r is:

  (x - h)² + (y - k)² = r²

- The center is equidistant from any point lies on the circumference

 of the circle

- There are three points equidistant from the center of the circle

- We have three unknowns in the equation of the circle h , k , r

- We will substitute the coordinates of these point in the equation of

 the circle to find h , k , r

* Lets solve the problem

∵ The equation of the circle is (x - h)² + (y - k)² = r²

∵ Points D (-2 , -4) , E (1 , 5) , F (2 , 0)

- Substitute the values of x and y b the coordinates of these points

# Point D (-2 , -4)

∵ (-2 - h)² + (-4 - k)² = r² ⇒ (1)

# Point E (1 , 5)

∵ (1 - h)² + (5 - k)² = r² ⇒ (2)

# Point (2 , 0)

∵ (2 - h)² + (0 - k)² = r²

∴ (2 - h)² + k² = r² ⇒ (3)

- To find h , k equate equation (1) , (2) and equation (2) , (3) because

  all of them equal r²

∵ (-2 - h)² + (-4 - k)² = (1 - h)² + (5 - k)² ⇒ (4)

∵ (1 - h)² + (5 - k)² = (2 - h)² + k² ⇒ (5)

- Simplify (4) and (5) by solve the brackets power 2

# (a ± b)² = (a)² ± (2 × a × b) + (b)²

# Equation (4)

∴ [(-2)² - (2 × 2 × h) + (-h)²] + [(-4)² - (2 × 4 × k) + (-k)²] =

  [(1)² - (2 × 1 × h) + (-h)²] + [(5)² - (2 × 5 × k) + (-k)²]

∴ 4 - 4h + h² + 16 - 8k + k² = 1 - 2h + h² + 25 - 10k + k² ⇒ add like terms

∴ 20 - 4h - 8k + h² + k² = 26 - 2h - 10k + h² + k² ⇒ subtract h² and k²

  from both sides

∴ 20 - 4h - 8k = 26 - 2h - 10k ⇒ subtract 20 and add 2h , 10k

  for both sides

∴ -2h + 2k = 6 ⇒ (6)

- Do the same with equation (5)

# Equation (5)

∴ [(1)² - (2 × 1 × h) + (-h)²] + [(5)² - (2 × 5 × k) + (-k)²] =

  [(2)² - (2 × 2 × h) + k²

∴ 1 - 2h + h² + 25 - 10k + k² = 4 - 4h + k²⇒ add like terms

∴ 26 - 2h - 10k + h² + k² = 4 - 4h + k² ⇒ subtract h² and k²

  from both sides

∴ 26 - 2h - 10k = 4 - 4h  ⇒ subtract 26 and add 4h

  for both sides

∴ 2h - 10k = -22 ⇒ (7)

- Add (6) and (7) to eliminate h and find k

∴ - 8k = -16 ⇒ divide both sides by -8

∴ k = 2

- Substitute this value of k in (6) or (7)

∴ 2h - 10(2) = -22

∴ 2h - 20 = -22 ⇒ add 20 to both sides

∴ 2h = -2 ⇒ divide both sides by 2

∴ h = -1

* The coordinates for the location of the center of the platform are (-1 , 2)