Find the equation of a circumcircle of a triangle whose testices are A (2, 3), B (5, 4) and C (3, 7)
I need the working and answer
To find the equation of the circumcircle of a triangle, we need to find the center and radius of the circle. The center of the circumcircle is the intersection of the perpendicular bisectors of any two sides of the triangle.
Step 1: Find the midpoints of two sides
We start by finding the midpoint of sides AB and BC.
Midpoint of AB:
x-coordinate = (x-coordinate of A + x-coordinate of B) / 2
y-coordinate = (y-coordinate of A + y-coordinate of B) / 2
Midpoint of AB:
x-coordinate = (2 + 5) / 2 = 7 / 2 = 3.5
y-coordinate = (3 + 4) / 2 = 7 / 2 = 3.5
Midpoint of BC:
x-coordinate = (x-coordinate of B + x-coordinate of C) / 2
y-coordinate = (y-coordinate of B + y-coordinate of C) / 2
Midpoint of BC:
x-coordinate = (5 + 3) / 2 = 8 / 2 = 4
y-coordinate = (4 + 7) / 2 = 11 / 2 = 5.5
So, the midpoint of side AB is (3.5, 3.5), and the midpoint of side BC is (4, 5.5).
Step 2: Find the slopes of the perpendicular bisectors
To find the equation of the perpendicular bisectors, we need the slope of the line passing through the midpoints of AB and BC (m1 and m2, respectively).
Slope of AB = (y-coordinate of B - y-coordinate of A) / (x-coordinate of B - x-coordinate of A)
Slope of AB = (4 - 3) / (5 - 2) = 1 / 3
Slope of BC = (y-coordinate of C - y-coordinate of B) / (x-coordinate of C - x-coordinate of B)
Slope of BC = (7 - 4) / (3 - 5) = 3 / -2 = -3 / 2
Step 3: Find the equation of the perpendicular bisectors
Using the point-slope form, we can write the equations of the perpendicular bisectors passing through the midpoints of AB and BC, respectively.
Equation of AB perpendicular bisector:
y - y1 = m1(x - x1)
y - 3.5 = (1/3)(x - 3.5)
y - 3.5 = (1/3)x - 7/2
y = (1/3)x - 7/2 + 7/2
y = (1/3)x
Equation of BC perpendicular bisector:
y - y2 = m2(x - x2)
y - 5.5 = (-3/2)(x - 4)
y - 5.5 = (-3/2)x + 6
y = (-3/2)x + 6 + 5.5
y = (-3/2)x + 11.5
Step 4: Find the intersection of the perpendicular bisectors
To find the center of the circumcircle, we solve the system of equations formed by the perpendicular bisectors. In this case, we need to find the intersection of the lines:
y = (1/3)x
y = (-3/2)x + 11.5
We can solve for x by equating the right-hand sides and solve for y using either equation. Let's solve using y = (1/3)x:
(1/3)x = (-3/2)x + 11.5
(1/3)x + (3/2)x = 11.5
(5/6)x = 11.5
x = (6/5)(11.5)
x = 13.8
Substituting this value back into y = (1/3)x:
y = (1/3)(13.8)
y = 4.6
So, the center of the circumcircle is (13.8, 4.6).
Step 5: Find the radius of the circumcircle
The distance between the center of the circumcircle and any vertex of the triangle is the radius of the circumcircle. Let's calculate the radius using the distance formula between (13.8, 4.6) and vertex A (2, 3):
r = √[(x2 - x1)^2 + (y2 - y1)^2]
r = √[(2 - 13.8)^2 + (3 - 4.6)^2]
r = √[(11.8)^2 + (-1.6)^2]
r = √[139.24 + 2.56]
r = √141.8
r ≈ 11.91
So, the radius of the circumcircle is approximately 11.91.
Step 6: Write the equation of the circumcircle
Finally, we can write the equation of the circumcircle using the center (h, k) and radius r in the standard form:
(x - h)^2 + (y - k)^2 = r^2
Plugging in the values:
(x - 13.8)^2 + (y - 4.6)^2 = (11.91)^2
Therefore, the equation of the circumcircle of the triangle with vertices A (2, 3), B (5, 4), and C (3, 7) is (x - 13.8)^2 + (y - 4.6)^2 = 141.8.
that's vertices, not testices.
The circumcenter lies at the intersection of the sides' perpendicular bisectors.
So, to start, find the equations of two of the bisectors.
AB: slope=1/3 midpoint is (7/2,7/2)
pb: (y-7/2) = (-3)(x-7/2)
AC: slope=4 midpoint is (5/2,5)
pb: (y-5) = (-1/4)(x-5/2)
the pb intersect at D:(67/22,107/22)
The distance AD is √2210/22
So, the circle is
(x-67/22)^2 + (y-107/22)^2 = 2210/484
see the graphs at
http://www.wolframalpha.com/input/?i=plot+y-3+%3D+%281%2F3%29%28x-2%29%2C+y-3+%3D+%284%29%28x-2%29%2C+y-4+%3D+%28-3%2F2%29%28x-5%29%2C+%28x-67%2F22%29^2+%2B+%28y-107%2F22%29^2+%3D+2210%2F484