Question

The equation of the circle passing through the points (1,2), (4,3), and (2,–1) is:

• A. \( x^2 + y^2 - 6x + 2y + 5 = 0 \)
• B. \( x^2 + y^2 - 7x + 4y + 6 = 0 \)
• C. \( x^2 + y^2 - 5x + 2y + 3 = 0 \)
• D. \( x^2 + y^2 - 6x + 2y + 6 = 0 \)

Hint:

Key Fact: Three non-collinear points determine a unique circle.

Answer 1

The Correct Option is A

The general equation of a circle is \(x^2+y^2+Dx+Ey+F=0\). To find the equation of the circle passing through (1,2), (4,3), and (2,–1), we substitute these points into the general equation to form a system of linear equations.
Step 1: Substituting (1,2) yields \(1^2+2^2+D(1)+E(2)+F=0\), which simplifies to \(D+2E+F=-5\).
Step 2: Substituting (4,3) yields \(4^2+3^2+D(4)+E(3)+F=0\), which simplifies to \(4D+3E+F=-25\).
Step 3: Substituting (2,-1) yields \(2^2+(-1)^2+D(2)+E(-1)+F=0\), which simplifies to \(2D-E+F=-5\).
The system of equations is:
\(D+2E+F=-5\)
\(4D+3E+F=-25\)
\(2D-E+F=-5\)
Step 4: Solve the system. Subtracting the first equation from the second yields \(3D+E=-20\). Subtracting the first equation from the third yields \(D-3E=0\). From \(D-3E=0\), we get \(D=3E\). Substituting this into \(3D+E=-20\) gives \(3(3E)+E=-20\), so \(10E=-20\), which means \(E=-2\). Substituting \(E=-2\) back into \(D=3E\) gives \(D=3(-2)=-6\). Finally, substituting \(D=-6\) and \(E=-2\) into \(D+2E+F=-5\) gives \(-6+2(-2)+F=-5\), so \(-6-4+F=-5\), which means \(F=5\).
The equation of the circle is \(x^2+y^2-6x+2y+5=0\).