Question:

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

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Key Fact: Three non-collinear points determine a unique circle.
Updated On: Jun 21, 2025
  • \( x^2 + y^2 - 6x + 2y + 5 = 0 \)
  • \( x^2 + y^2 - 7x + 4y + 6 = 0 \)
  • \( x^2 + y^2 - 5x + 2y + 3 = 0 \)
  • \( x^2 + y^2 - 6x + 2y + 6 = 0 \)
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The Correct Option is A

Approach Solution - 1

To find the equation of the circle passing through the given points (1,2), (4,3), and (2,–1), we start with the general form of the equation of a circle:
\(x^2+y^2+Dx+Ey+F=0\)
Substituting the given points into this equation gives us a system of linear equations. 
Step 1: Substitute (1,2) into the circle equation:
\(1^2+2^2+D(1)+E(2)+F=0\)
\(1+4+D+2E+F=0\)
\(D+2E+F=-5\)
Step 2: Substitute (4,3) into the circle equation:
\(4^2+3^2+D(4)+E(3)+F=0\)
\(16+9+4D+3E+F=0\)
\(4D+3E+F=-25\)
Step 3: Substitute (2,-1) into the circle equation:
\(2^2+(-1)^2+D(2)+E(-1)+F=0\)
\(4+1+2D-E+F=0\)
\(2D-E+F=-5\)
We now have a system of equations:
\(D+2E+F=-5\)
\(4D+3E+F=-25\)
\(2D-E+F=-5\)
Step 4: Solve these linear equations using elimination or substitution. Solve for \(F\) using any two of the equations. Subtract the first equation from the second:
\(3D+E=-20\)
Subtract the first equation from the third:
\(D-3E=0\)
Solve the two new equations:
From \(D-3E=0\), we have \(D=3E\). Substitute into \(3D+E=-20\):
\(3(3E)+E=-20\)
\(9E+E=-20\)
\(10E=-20\)
\(E=-2\)
Substitute \(E=-2\) back into \(D=3E\):
\(D=3(-2)=-6\)
Finally, from \(D+2E+F=-5\), we substitute \(D=-6\) and \(E=-2\):
\(-6+2(-2)+F=-5\)
\(-6-4+F=-5\)
\(F=5\)
Thus, the equation of the circle is:
\(x^2+y^2-6x+2y+5=0\)

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Approach Solution -2

To find the equation of the circle passing through three points, we use the general form of a circle equation:

\(x^2+y^2+Dx+Ey+F=0\)

We need to determine \(D\), \(E\), and \(F\) such that the circle passes through the given points \((1,2)\), \((4,3)\), and \((2,-1)\).

Substituting each point into the circle equation:

For \((1,2)\):

\(1^2+2^2+D\cdot1+E\cdot2+F=0\) simplifies to \(1+4+D+2E+F=0\). Thus, we have equation (1):

\(D+2E+F=-5\)

For \((4,3)\):

\(4^2+3^2+D\cdot4+E\cdot3+F=0\) simplifies to \(16+9+4D+3E+F=0\). Thus, we have equation (2):

\(4D+3E+F=-25\)

For \((2,-1)\):

\(2^2+(-1)^2+D\cdot2+E\cdot(-1)+F=0\) simplifies to \(4+1+2D-E+F=0\). Thus, we have equation (3):

\(2D-E+F=-5\)

We solve the system of equations:

1. \(D+2E+F=-5\)

2. \(4D+3E+F=-25\)

3. \(2D-E+F=-5\)

Subtract equation (1) from equation (2):

\((4D+3E+F)-(D+2E+F)=-25+5\)

\(3D+E=-20\)   (Equation 4)

Subtract equation (3) from equation (1):

\((D+2E+F)-(2D-E+F)=-5+5\)

\(-D+3E=0\)   (Equation 5)

From equation (5), we have:

\(D=3E\)

Substitute \(D=3E\) into equation (4):

\(3(3E)+E=-20\)

\(9E+E=-20\)

\(10E=-20\)

\(E=-2\)

Substitute \(E=-2\) back into \(D=3E\):

\(D=3(-2)=-6\)

Substitute \(D=-6\) and \(E=-2\) into equation (1):

\(-6+2(-2)+F=-5\)

\(-6-4+F=-5\)

\(F=5\)

Thus, the equation of the circle is:

\(x^2+y^2-6x+2y+5=0\)

This matches the first option, confirming it as the correct solution.

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