Math Problem Statement

Find the equation of circle passing through the vertices of triangle whose sides are represented by the equation x+y=2, 3x-4y=6, x-y=0

Solution

To find the equation of the circle passing through the vertices of the triangle formed by the given lines, follow these steps:

Step 1: Find the Vertices of the Triangle

The given lines are:

  1. x+y=2x + y = 2
  2. 3x4y=63x - 4y = 6
  3. xy=0x - y = 0 (or x=yx = y)

Find the intersection points of these lines:

Vertex A: Intersection of x+y=2x + y = 2 and xy=0x - y = 0

Solving simultaneously: x+y=2x + y = 2 xy=0x=yx - y = 0 \Rightarrow x = y Substituting x=yx = y in x+y=2x + y = 2: x+x=22x=2x=1,y=1x + x = 2 \Rightarrow 2x = 2 \Rightarrow x = 1, y = 1 So, A(1,1)A(1,1).

Vertex B: Intersection of xy=0x - y = 0 and 3x4y=63x - 4y = 6

Substituting x=yx = y in 3x4y=63x - 4y = 6: 3y4y=6y=6y=63y - 4y = 6 \Rightarrow -y = 6 \Rightarrow y = -6 Since x=yx = y, we get x=6x = -6.
So, B(6,6)B(-6,-6).

Vertex C: Intersection of x+y=2x + y = 2 and 3x4y=63x - 4y = 6

Multiply x+y=2x + y = 2 by 3: 3x+3y=63x + 3y = 6 Now, subtract from 3x4y=63x - 4y = 6: (3x4y)(3x+3y)=66(3x - 4y) - (3x + 3y) = 6 - 6 4y3y=07y=0y=0-4y - 3y = 0 \Rightarrow -7y = 0 \Rightarrow y = 0 Substituting y=0y = 0 in x+y=2x + y = 2: x+0=2x=2x + 0 = 2 \Rightarrow x = 2 So, C(2,0)C(2,0).


Step 2: Find the Equation of the Circumcircle

The general equation of a circle is: x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 Substituting the three points A(1,1)A(1,1), B(6,6)B(-6,-6), and C(2,0)C(2,0) into the equation:

Using Point A (1,1):

12+12+D(1)+E(1)+F=01^2 + 1^2 + D(1) + E(1) + F = 0 1+1+D+E+F=01 + 1 + D + E + F = 0 D+E+F=2D + E + F = -2

Using Point B (-6,-6):

(6)2+(6)2+D(6)+E(6)+F=0(-6)^2 + (-6)^2 + D(-6) + E(-6) + F = 0 36+366D6E+F=036 + 36 - 6D - 6E + F = 0 6D6E+F=72-6D - 6E + F = -72

Using Point C (2,0):

(2)2+(0)2+D(2)+E(0)+F=0(2)^2 + (0)^2 + D(2) + E(0) + F = 0 4+2D+F=04 + 2D + F = 0 2D+F=42D + F = -4


Step 3: Solve for D, E, and F

We have the system:

  1. D+E+F=2D + E + F = -2
  2. 6D6E+F=72-6D - 6E + F = -72
  3. 2D+F=42D + F = -4

Solve for FF from equation (3): F=42DF = -4 - 2D

Substituting in equation (1): D+E+(42D)=2D + E + (-4 - 2D) = -2 D+E42D=2D + E - 4 - 2D = -2 ED=2-E - D = 2 E+D=2(Equation 4)E + D = -2 \quad \text{(Equation 4)}

Substituting F=42DF = -4 - 2D in equation (2): 6D6E+(42D)=72-6D - 6E + (-4 - 2D) = -72 8D6E=68-8D - 6E = -68 4D+3E=34(Equation 5)4D + 3E = 34 \quad \text{(Equation 5)}

Solve equations (4) and (5):

Multiply (4) by 3: 3D+3E=63D + 3E = -6 Subtract from (5): (4D+3E)(3D+3E)=34(6)(4D + 3E) - (3D + 3E) = 34 - (-6) D=40D = 40

Substituting D=40D = 40 in (4): 40+E=240 + E = -2 E=42E = -42

Substituting D=40D = 40 in F=42DF = -4 - 2D: F=480=84F = -4 - 80 = -84


Step 4: Write the Equation of the Circle

x2+y2+40x42y84=0x^2 + y^2 + 40x - 42y - 84 = 0


Final Answer:

x2+y2+40x42y84=0\mathbf{x^2 + y^2 + 40x - 42y - 84 = 0}

Would you like me to simplify further or explain any step?

Here are 5 related problems you can try:

  1. Find the circumcenter of the triangle formed by the given three lines.
  2. Find the radius of the circumcircle of the triangle.
  3. Find the equation of the centroid of the triangle.
  4. Find the area of the triangle formed by the given three lines.
  5. Find the equation of the orthocenter of the triangle.

Tip:

To find the radius of the circumcircle, use the formula: R=a2+b2c22R = \frac{\sqrt{a^2 + b^2 - c^2}}{2} where a,b,ca, b, c are the distances between the vertices.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Circle Equations
Intersection of Lines
Circumcircle

Formulas

General equation of a circle: x^2 + y^2 + Dx + Ey + F = 0
Intersection of two lines (solving system of equations)

Theorems

Circumcenter of a triangle
Equation of Circumcircle

Suitable Grade Level

Grades 11-12