Find the equation of a circle circumscribes a triangle determined by the line y= 0 , y= x and 2x+3y= 10
PLEASE HELP ME BELLS
Make a sketch, and by some simple preliminary algebra you can see that the triangle is
A(0,0) , B(2,2) and C(0, 10/3)
the centre of the circle lies on the right bisectors of any of these sides
midpoint of AC = (0, 5/3) so right-bisector equation is
y = 5/3
midpoint of B(2,2) is (1,1)
slope of AB = 1
so slope of bisector at (1,1) is =1
y-1 = -1(x-1)
x + y = 2
sub in y = 5/3
x = 2-5/3 = 1/3
centre is (1/3 , 5/3)
equation:
(x - 1/3)^2 + (y-5/3)^2 = r^2
plug in (0,0) which lies on it
1/9 + 25/9 = r^2 = 26/9
(x-1/3)^2 + (y-5/3)^2 = 26/9
oops, I misread the question, and got the wrong triangle
(I used the y-axis, instead of the x-axis)
so thepoints are (0,0), (2,2) and (5,0)
same equation for the right-bisector of AB which is
x + y = 2
the right bisector of line from A to (5,0) is
x = 5/2
for centre:
5/2 + y = 2
y = -1/2
centre is (5/2, -1/2)
(x-5/2)^2 + (y+1/2)^2 = r^2
for (0,0) , which lies on the circle
25/4 + 1/4 = r^2 = 26/4
(x - 5/2)^2 + (y+ 1/2)^2 = 26/4
To find the equation of a circle that circumscribes a triangle, you need to find the coordinates of the triangle's vertices first. In this case, the triangle is determined by the lines y = 0, y = x, and 2x + 3y = 10.
1. Find the coordinates of the intersection points:
To find the first vertex, set y = 0 in the equation y = x. This gives us x = 0, so the first vertex is (0, 0).
To find the second vertex, determine the intersection point of the lines y = x and 2x + 3y = 10. Substitute y = x into the second equation to get 2x + 3x = 10, which simplifies to 5x = 10. Solving for x gives x = 2, and substituting this value back into the equation y = x gives y = 2. So the second vertex is (2, 2).
2. Find the length of the line segments that form the triangle:
The two sides of the triangle are the lines y = 0 and y = x. The length of the line segment between the first and second vertices can be found using the distance formula. The distance between (0, 0) and (2, 2) is given by:
√((2 - 0)^2 + (2 - 0)^2) = √(4 + 4) = √8 = 2√2.
So the length of this line segment is 2√2.
3. Find the coordinates of the center of the circle:
The center of the circle that circumscribes a triangle can be found by finding the intersection of the perpendicular bisectors of any two sides of the triangle. We will find the intersection of the perpendicular bisectors of the lines y = 0 and y = x.
- Perpendicular bisector of y = 0:
The line y = 0 is a horizontal line. The perpendicular bisector of a horizontal line is a vertical line passing through the midpoint of the line segment. The midpoint of the line segment between (0, 0) and (2, 0) is ((0 + 2)/2, (0 + 0)/2) = (1, 0).
- Perpendicular bisector of y = x:
The line y = x has a slope of 1, so the perpendicular bisector will have a slope of -1 (the negative reciprocal). The midpoint of the line segment between (0, 0) and (2, 2) is ((0 + 2)/2, (0 + 2)/2) = (1, 1). Using the point-slope form, we get the equation of the perpendicular bisector as:
y - 1 = -1(x - 1)
y - 1 = -x + 1
y = -x + 2
Setting these two equations for the perpendicular bisectors equal to each other gives:
-x + 2 = 0
x = 2.
Substituting this back into either equation, we get y = 0.
Therefore, the center of the circle is (2, 0).
4. Find the radius of the circle:
The radius of the circle is the distance between the center and any of the vertices of the triangle. Let's use the first vertex (0, 0).
The distance between (2, 0) and (0, 0) is given by:
√((2 - 0)^2 + (0 - 0)^2) = √(4 + 0) = √4 = 2.
So the radius of the circle is 2.
5. Write the equation of the circle:
The equation of a circle with center (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2.
Substituting the values we found, the equation of the circle is:
(x - 2)^2 + (y - 0)^2 = 2^2.
Simplifying, the equation of the circle is:
(x - 2)^2 + y^2 = 4.
So, the equation of the circle that circumscribes the triangle determined by the lines y = 0, y = x, and 2x + 3y = 10 is (x - 2)^2 + y^2 = 4.