Question

Equations of the circle circumscribing the triangle formed by the lines \(x + y = 6\), \(2x + y = 4\), \(x + 2y = 5\) is:

• A. \(x^2 + y^2 - 17x + 50y - 19 = 0\)
• B. \(x^2 + y^2 + 17x + 19y - 25 = 0\)
• C. \(x^2 + y^2 + 17x - 19y + 50 = 0\)
• D. \(x^2 + y^2 - 17x - 19y + 50 = 0\)

Hint:

Exam Tip:
For circumcircles of triangles:

• Find the vertices.
• Use the general equation of a circle.
• Solve the system of equations for D, E, F.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to find the equation of the circumcircle of a triangle formed by three lines. The circle passes through the three vertices of the triangle.

Step 2: Key Formula or Approach:
Find the vertices by solving the pairs of equations. Then use the general equation of a circle \(x^2 + y^2 + Dx + Ey + F = 0\) and substitute the three vertices to solve for D, E, F.

Step 3: Detailed Explanation:
Find the vertices:
• Intersection of \(x + y = 6\) and \(2x + y = 4\):
Subtract: \(2x + y - (x + y) = 4 - 6 \Rightarrow x = -2\).
Then \(-2 + y = 6 \Rightarrow y = 8\). So, vertex A = \((-2, 8)\).
• Intersection of \(x + y = 6\) and \(x + 2y = 5\):
Subtract: \(x + 2y - (x + y) = 5 - 6 \Rightarrow y = -1\).
Then \(x - 1 = 6 \Rightarrow x = 7\). So, vertex B = \((7, -1)\).
• Intersection of \(2x + y = 4\) and \(x + 2y = 5\):
Multiply second by 2: \(2x + 4y = 10\). Subtract first: \(3y = 6 \Rightarrow y = 2\).
Then \(x + 4 = 5 \Rightarrow x = 1\). So, vertex C = \((1, 2)\). Now, let the circle be \(x^2 + y^2 + Dx + Ey + F = 0\).
Substitute the three points:
• For A \((-2, 8)\): \(4 + 64 - 2D + 8E + F = 0 \Rightarrow 68 - 2D + 8E + F = 0\).
• For B \((7, -1)\): \(49 + 1 + 7D - E + F = 0 \Rightarrow 50 + 7D - E + F = 0\).
• For C \((1, 2)\): \(1 + 4 + D + 2E + F = 0 \Rightarrow 5 + D + 2E + F = 0\). Subtract the third equation from the first: \[ (68 - 2D + 8E + F) - (5 + D + 2E + F) = 0 \Rightarrow 63 - 3D + 6E = 0 \Rightarrow -3D + 6E = -63 \Rightarrow D - 2E = 21 \quad \text{(1)} \] Subtract the third equation from the second: \[ (50 + 7D - E + F) - (5 + D + 2E + F) = 0 \Rightarrow 45 + 6D - 3E = 0 \Rightarrow 6D - 3E = -45 \Rightarrow 2D - E = -15 \quad \text{(2)} \] From (1): \(D = 21 + 2E\).
Substitute into (2): \(2(21 + 2E) - E = -15 \Rightarrow 42 + 4E - E = -15 \Rightarrow 3E = -57 \Rightarrow E = -19\).
Then \(D = 21 + 2(-19) = 21 - 38 = -17\).
Now, from the third equation: \(5 + D + 2E + F = 0 \Rightarrow 5 - 17 - 38 + F = 0 \Rightarrow -50 + F = 0 \Rightarrow F = 50\).
So, the circle is: \(x^2 + y^2 - 17x - 19y + 50 = 0\).
This matches option (D).

Step 4: Final Answer:
Therefore, option (D) is correct.