Asked by Kim Jisoo
Find the center of the circle with passing length through the points A(1,2) , B(3,7) , C(5,3).
pl answer it's urgent :)
pl answer it's urgent :)
Answers
Answered by
oobleck
find where the perpendicular bisectors of the segments AB and AC intersect. That is the center of the circle.
AB has slope 5/2 and its midpoint is (2,9/2)
AC has slope 1/4 and its midpoint is (3,5/2)
So you just have to solve
y-9/2 = -2/5 (x-2)
y-5/2 = -4(x-3)
so the center is at (23/9,77/18)
That makes the radius √((23/9-1)^2+(77/18-2)^2) = √2465/18
so the equation is
(x - 23/9)^2 + (y - 77/18)^2 = 2465/324
AB has slope 5/2 and its midpoint is (2,9/2)
AC has slope 1/4 and its midpoint is (3,5/2)
So you just have to solve
y-9/2 = -2/5 (x-2)
y-5/2 = -4(x-3)
so the center is at (23/9,77/18)
That makes the radius √((23/9-1)^2+(77/18-2)^2) = √2465/18
so the equation is
(x - 23/9)^2 + (y - 77/18)^2 = 2465/324
Answered by
Bosnian
OR
The general form of the equation of the circle with center ( x₀ , y₀ ) and radius R is:
x² + y² + A x + B y + C = 0
where:
A = - 2 x₀
B = - 2 y₀
C = x₀² + y₀² - R²
In this case:
First point:
x = 1 , y = 2
x² + y² + A x + B y + C = 0
1² + 2² + A ∙ 1 + B ∙ 2 + C = 0
1 + 4 + A + 2 B + C = 0
5 + A + 2 B + C = 0
A + 2 B + C = - 5
Second point:
x = 3 , y = 7
x² + y² + A x + B y + C = 0
3² + 7² + A ∙ 3 + B ∙ 7 + C = 0
9 + 49 + 3 A + 7 B + C = 0
58 + 3 A + 7 B + C = 0
3 A + 7 B + C = - 58
Third point:
x = 5 , y = 3
x² + y² + A x + B y + C = 0
5² + 3² + A ∙ 5 + B ∙ 3 + C = 0
25 + 9 + 5 A + 3 B + C = 0
34 + 5 A + 3 B + C = 0
5 A + 3 B + C = - 34
Solve system:
A + 2 B + C = - 5
3 A + 7 B + C = - 58
5 A + 3 B + C = - 34
________________
The solutions are:
A = - 46 / 9 , B = - 77 / 9 , C = 155 / 9
Now:
A = - 2 x1
- 46 / 9 = - 2 x₀
x₀ = 23 / 9
B = - 2 y₀
- 77 / 9 = - 2 y₀
y₀ = 77 / 18
C = x₀² + y₀² - R²
155 / 9 = ( 23 / 9 )² + ( 77 / 18 )² - R²
R² = 2465 / 324
Now you can write equation of a circle in standard form:
( x - x₀ )² + ( y - y₀ )² = R²
( x - 23 / 9 )² + ( y - 77 / 18 )² = 2465 / 324
The general form of the equation of the circle with center ( x₀ , y₀ ) and radius R is:
x² + y² + A x + B y + C = 0
where:
A = - 2 x₀
B = - 2 y₀
C = x₀² + y₀² - R²
In this case:
First point:
x = 1 , y = 2
x² + y² + A x + B y + C = 0
1² + 2² + A ∙ 1 + B ∙ 2 + C = 0
1 + 4 + A + 2 B + C = 0
5 + A + 2 B + C = 0
A + 2 B + C = - 5
Second point:
x = 3 , y = 7
x² + y² + A x + B y + C = 0
3² + 7² + A ∙ 3 + B ∙ 7 + C = 0
9 + 49 + 3 A + 7 B + C = 0
58 + 3 A + 7 B + C = 0
3 A + 7 B + C = - 58
Third point:
x = 5 , y = 3
x² + y² + A x + B y + C = 0
5² + 3² + A ∙ 5 + B ∙ 3 + C = 0
25 + 9 + 5 A + 3 B + C = 0
34 + 5 A + 3 B + C = 0
5 A + 3 B + C = - 34
Solve system:
A + 2 B + C = - 5
3 A + 7 B + C = - 58
5 A + 3 B + C = - 34
________________
The solutions are:
A = - 46 / 9 , B = - 77 / 9 , C = 155 / 9
Now:
A = - 2 x1
- 46 / 9 = - 2 x₀
x₀ = 23 / 9
B = - 2 y₀
- 77 / 9 = - 2 y₀
y₀ = 77 / 18
C = x₀² + y₀² - R²
155 / 9 = ( 23 / 9 )² + ( 77 / 18 )² - R²
R² = 2465 / 324
Now you can write equation of a circle in standard form:
( x - x₀ )² + ( y - y₀ )² = R²
( x - 23 / 9 )² + ( y - 77 / 18 )² = 2465 / 324
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