Question

Find the equation of the circle which passes through the points $ (1,2) $, $ (4,3) $ and has its center on the line $ x + y = 5 $.

• A. \( (x - 2)^2 + (y - 3)^2 = 5 \)
• B. \( (x - 3)^2 + (y - 2)^2 = 2 \)
• C. \( (x - 2.5)^2 + (y - 2.5)^2 = 2.5 \)
• D. \( (x - 2)^2 + (y - 3)^2 = 2 \)

Hint:

Tip: Use system of equations and the condition for equal radii distances from the center to points on the circle.

Answer 1

The Correct Option is C

Step 1: Assume center coordinates 
Let the center be \( (h, k) \) lying on the line \( x + y = 5 \), so \[ h + k = 5 \quad \Rightarrow \quad k = 5 - h \]

Step 2: Use the fact that both points lie on the circle 
The radius squared \( r^2 = (1 - h)^2 + (2 - k)^2 = (4 - h)^2 + (3 - k)^2 \).

Step 3: Set the distances equal 
\[ (1 - h)^2 + (2 - k)^2 = (4 - h)^2 + (3 - k)^2 \] Expand: \[ (1 - h)^2 = (h - 1)^2 = h^2 - 2h + 1 \] \[ (2 - k)^2 = (k - 2)^2 = k^2 - 4k + 4 \] \[ (4 - h)^2 = (h - 4)^2 = h^2 - 8h + 16 \] \[ (3 - k)^2 = (k - 3)^2 = k^2 - 6k + 9 \] Substitute: \[ h^2 - 2h + 1 + k^2 - 4k + 4 = h^2 - 8h + 16 + k^2 - 6k + 9 \]

Step 4: Simplify the equation 
Cancel \( h^2 \) and \( k^2 \) on both sides: \[ -2h + 1 - 4k + 4 = -8h + 16 - 6k + 9 \] \[ (-2h - 4k + 5) = (-8h - 6k + 25) \] Rearrange: \[ -2h - 4k + 5 + 8h + 6k - 25 = 0 \implies 6h + 2k - 20 = 0 \] Divide by 2: \[ 3h + k = 10 \]

Step 5: Use the line condition 
Recall from Step 1: \[ k = 5 - h \] Substitute in \( 3h + k = 10 \): \[ 3h + (5 - h) = 10 \Rightarrow 2h + 5 = 10 \Rightarrow 2h = 5 \Rightarrow h = \frac{5}{2} = 2.5 \] Then: \[ k = 5 - 2.5 = 2.5 \]

Step 6: Calculate radius squared \( r^2 \) 
Using point \( (1, 2) \): \[ r^2 = (1 - 2.5)^2 + (2 - 2.5)^2 = (-1.5)^2 + (-0.5)^2 = 2.25 + 0.25 = 2.5 \]

Step 7: Final equation 
The circle equation is: \[ (x - 2.5)^2 + (y - 2.5)^2 = 2.5 \]