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

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

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)

Answer 2

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

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)