Find the center of the circle that can be circumscribed about triangle ABC with A(0, 0), B(6, 0), C(6, 4).

(6, 2)
(3, 4)
(2, 3)
(3, 2)

Respuesta :

Answer:

(3,2) is the center of the circle that can be circumscribed about triangle ABC with A(0, 0), B(6, 0), C(6, 4).

Step-by-step explanation:

The standard equation of a circle with center (h, k) and radius "r" is given by:

(x - h)²+ (y - k)² = r²

By using the 3 unknown points on the circle, we can have three equations that have to be solved simultaneously.

So,

Using point A(0, 0) in standard equation of a circle:

(x - h)²+ (y - k)² = r²

(0 - h)²+ (0 - k)² = r²

h² + k² = r²     Equation ( 1 )

Using point B(6, 0) in standard equation of a circle:

(x - h)²+ (y - k)² = r²

(6 - h)²+ (0 - k)² = r²

36 - 12h + h² + k² = r²      Equation ( 2 )

Using point B(6, 4) in standard equation of a circle:

(x - h)²+ (y - k)² = r²

(6 - h)²+ (4 - k)² = r²

36 - 12h + h² + 16 - 8K + k² = r²

52 - 12h + h² - 8K + k² = r²      Equation ( 3 )

Subtracting  Equation ( 1 ) from Equation ( 2 )

36 - 12h + h² + k² = r²

               - h² ± k² = - r²

···············································

36 - 12h = 0  Equation ( A )

···············································

Subtracting  Equation ( 1 ) from Equation ( 3 )

52 - 12h + h² - 8k + k² = r²  

              - h²          ± k²= - r²

···············································

52 - 12h - 8k = 0  Equation ( B )

···············································    

By Equation (A)

36 - 12h = 0

12h = 36

h = 3

Putting the equation h = 3 in Equation (B)

52 - 12h - 8k = 0

52 - 12(3) - 8k = 0

52 - 36 = 8k

16 = 8k

k = 2

So, the center of circle = (h, k) = (3, 2)

Keywords: circle, center of circle, triangle

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