How to Find the Equation of a Circle Passing Through Two Points and Its Center on a Given Line

To find the equation of a circle that passes through two given points and whose center lies on a specific line, we can follow a systematic approach. This article will walk you through the process step-by-step, providing a detailed solution to the problem of finding the equation of the circle passing through points A(-2, -1) and B(5, 1) with its center on the line 3x y - 2 0.

Step-by-Step Solution

Step 1: General Equation of the Circle

The general equation of a circle with center (h, k) and radius r is:

[ x^2 - 2hx h^2 y^2 - 2ky k^2 r^2 ]

Alternatively, it can be written as:

[ (x - h)^2 (y - k)^2 r^2 ]

Step 2: Equation of the Line for the Center

The center (h, k) lies on the line given by the equation 3x y - 2 0. This can be used to express k in terms of h:

[ k -3h - 2 ]

Step 3: Condition for the Circle to Pass Through Points

The circle must pass through the points A(-2, -1) and B(5, 1). This gives us two conditions:

For point A(-2, -1):

[ (-2 - h)^2 (-1 - k)^2 r^2 ]

For point B(5, 1):

[ (5 - h)^2 (1 - k)^2 r^2 ]

Step 4: Substitute k in the Circle Equations

Substitute ( k -3h - 2 ) into both equations:

For point A(-2, -1):

[ (-2 - h)^2 (-1 - (-3h - 2))^2 r^2 ]

Expanding and simplifying:

[ (-2 - h)^2 (2 3h)^2 r^2 ]

[ 4 4h h^2 4 12h 9h^2 r^2 ]

[ 10h^2 16h 8 r^2 quad text{(Equation 1)} ]

For point B(5, 1):

[ (5 - h)^2 (1 - (-3h - 2))^2 r^2 ]

Expanding and simplifying:

[ (5 - h)^2 (3 3h)^2 r^2 ]

[ 25 - 10h h^2 9 18h 9h^2 r^2 ]

[ 10h^2 8h 34 r^2 quad text{(Equation 2)} ]

Step 5: Set the Two Expressions for ( r^2 ) Equal

Equating the two expressions for ( r^2 ) gives:

[ 10h^2 16h 8 10h^2 8h 34 ]

Cancelling ( 10h^2 ) from both sides:

[ 16h 8 8h 34 ]

Subtracting ( 8h ) and 8 from both sides:

[ 8h 26 ]

Solving for ( h ):

[ h frac{26}{8} frac{13}{4} times frac{1}{2} frac{13}{8} ]

Step 6: Find ( k )

Substituting ( h frac{13}{8} ) into ( k -3h - 2 ):

[ k -3 left( frac{13}{8} right) - 2 ]

[ k -frac{39}{8} - frac{16}{8} -frac{55}{8} ]

Step 7: Find ( r^2 )

Substituting ( h frac{13}{8} ) and ( k -frac{55}{8} ) back into either equation for ( r^2 ). Using Equation 1:

[ r^2 10 left( frac{13}{8} right)^2 16 left( frac{13}{8} right) 8 ]

[ r^2 10 left( frac{169}{64} right) 16 left( frac{13}{8} right) 8 ]

[ r^2 frac{1690}{64} frac{208}{8} 8 ]

[ r^2 frac{1690}{64} frac{1664}{64} frac{512}{64} ]

[ r^2 frac{1690 1664 512}{64} frac{3866}{64} frac{1933}{32} ]

Step 8: Final Equation of the Circle

The center of the circle is ( left( frac{13}{8}, -frac{55}{8} right) ) and the radius squared is ( r^2 frac{1933}{32} ).

Hence, the equation of the circle is:

[ left(x - frac{13}{8}right)^2 left(y frac{55}{8}right)^2 frac{1933}{32} ]

This can be rewritten in standard form or left as is for clarity.

/p

The final answer is: (left(x - frac{13}{8}right)^2 left(y frac{55}{8}right)^2 frac{1933}{32})