Question

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 :)

Answer 1

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

Answer 2

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