Question

The three vertices of a triangle are \( (0,0), (3,1) \) and \( (1,3) \). If this triangle is inscribed in a circle, then the equation of the circle is:

• A. \( 2x^2 + 2y^2 - 2x - 6y = 0 \)
• B. \( x^2 + y^2 - 3x - y = 0 \)
• C. \( x^2 + y^2 - x - 3y = 0 \)
• D. \( 2x^2 + 2y^2 - 6x - 2y = 0 \)
• E. \( 2x^2 + 2y^2 - 5x - 5y = 0 \)

Hint:

To find the equation of the circumcircle, use the general form \( x^2 + y^2 + Dx + Ey + F = 0 \) and substitute the coordinates of the triangle's vertices to solve for the constants \( D \), \( E \), and \( F \).

Answer 1

Answer: (E)

Step 1:Understand the condition for the equation of the circumcircle.
The equation of the circumcircle of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) can be written in the general form: \[ x^2 + y^2 + Dx + Ey + F = 0 \] where \( D, E, F \) are constants that can be determined using the coordinates of the vertices.
Step 2:Set up the system of equations.
The given vertices are \( (0, 0), (3, 1), (1, 3) \). Substituting these into the general equation \( x^2 + y^2 + Dx + Ey + F = 0 \), we get the following system of equations: 1. For \( (0, 0) \), we substitute \( x = 0 \) and \( y = 0 \): \[ 0 + 0 + D(0) + E(0) + F = 0 \quad \Rightarrow \quad F = 0 \] 2. For \( (3, 1) \), we substitute \( x = 3 \) and \( y = 1 \): \[ 3^2 + 1^2 + D(C) + E(A) + 0 = 0 \quad \Rightarrow \quad 9 + 1 + 3D + E = 0 \] Simplifying: \[ 10 + 3D + E = 0 \quad \Rightarrow \quad 3D + E = -10 \tag{1} \] 3. For \( (1, 3) \), we substitute \( x = 1 \) and \( y = 3 \): \[ 1^2 + 3^2 + D(A) + E(C) + 0 = 0 \quad \Rightarrow \quad 1 + 9 + D + 3E = 0 \] Simplifying: \[ 10 + D + 3E = 0 \quad \Rightarrow \quad D + 3E = -10 \tag{2} \]
Step 3:Solve the system of equations.
From equations (A) and (B): 1. \( 3D + E = -10 \) 2. \( D + 3E = -10 \) Multiply equation (B) by 3: \[ 3D + 9E = -30 \] Now subtract equation (A) from this result: \[ (3D + 9E) - (3D + E) = -30 - (-10) \] \[ 8E = -20 \quad \Rightarrow \quad E = -\frac{5}{2} \] Substitute \( E = -\frac{5}{2} \) into equation (A): \[ 3D - \frac{5}{2} = -10 \quad \Rightarrow \quad 3D = -10 + \frac{5}{2} = -\frac{15}{2} \] \[ D = -\frac{5}{2} \]
Step 4:Substitute \( D \) and \( E \) back into the general equation.
We now substitute \( D = -\frac{5}{2} \) and \( E = -\frac{5}{2} \) into the equation \( x^2 + y^2 + Dx + Ey + F = 0 \): \[ x^2 + y^2 - \frac{5}{2}x - \frac{5}{2}y = 0 \] Multiply through by 2 to eliminate the fractions: \[ 2x^2 + 2y^2 - 5x - 5y = 0 \] This matches option (E).