Find the vector from the origin to the point of intersection of the medians of the triangle whose vertices are A(-1,-3),B(5,7), and C(2,5).
Answers
Find the circumcentre of the triangle with the vertices A (-3,4) , B (3,4) and C (-4,-3). What is the circumradius and area of the circle?
Find the circumcentre of the triangle with the vertices A (-3,4) , B (3,4) and C (-4,-3). What is the circumradius and area of the circle?
We want the equation of the circle which passes through the points (−3,4), (3,4), and (−4,3).
The equation of a circle is x2+y2+ax+by+c=0.
From the coordinates of the three given points we get following three equations:
(−3)2+42−3a+4b+c=0
32+42+3a+4b+c=0
(−4)2+32−4a+3b+c=0
We can simplify the above equations, and put them in matrix form:
⎡⎣⎢−33−4443111⎤⎦⎥⎡⎣⎢abc⎤⎦⎥=⎡⎣⎢−25−25−25⎤⎦⎥
Solving the above matrix equation using Cramer's method, we get a=0, b=0, c=−25.
Plugging in these values of a, b, and c into the general form of the equation, we can now write the equation of the required circle as
x2+y2−25=0,
Or, expressing it in center radius form,
(x+0)2+(y+0)2=25
The center of the circle is (0,0) and the radius of the circle is 5.
The area of a circle with radius 5 is πr2 = 25π.
6.1k Views ·
Letx2+y2+2gx+2fy+c=0 be the equation of circumcircle.
plugging in the coordinates of 3 points we get
-6g+8f+c=-25 …..(1)
6g+8f+c=-25 …..(2)
-8g-6f+c=-25 ……(3)
(2) - (1) gives 12g=0∴g=0
(2)-(3) gives 2g+14f=0 i.e. 14f=0 f=0
Circum center=(-g,-f)=(0,0)
Substituting values of g and f ,c=25 R=g2+f2−c−−−−−−−−−√=5
Area of circle=25π
4.1k Views
-A2A-
Circumcenter of a triangle is the intersection point of the perpendicular bisectors of the three sides.
Perpendicular bisector of a line segment is a line passing through the mid-point of the line segment and is perpendicular to it.
Perpendicular bisector of AB:
Mid-point of AB, M(−3+32, 4+42)
Coordinates of M(0, 4)
Gradient or slope of AB, m = 4−4−3−3 =0
Gradient or slope of line perpendicular to AB = −1m =∞
⇒ Perpendicular line to AB is a vertical line on xy plane.
Perpendicular bisector of AB is a vertical line passing through M(0,4). It's equation: x=0 ----------Line 1
Perpendicular bisector of BC:
Mid-point of BC, N(3+(−4)2, 4+(−3)2)
Coordinates of N(−12, 12)
Gradient or slope of BC, m = −3−4−4−3 =1
Gradient or slope of line perpendicular to BC = −1m =−1
Perpendicular bisector of BC is a line passing through N (−12, 12) and is having a slope -1.
Equation of perpendicular bisector of BC:
y−12 =−1(x−(−12))
y−12 =−x−12
y = x --------- Line 2
Circumcenter is point of intersection of Line 1 and Line 2.
x= 0, y =0
Therefore, circumcenter of the given triangle is the origin O(0,0)
I hope it helps!
