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

To find the equation of the circle that passes through the points \((1,2)\), \((4,3)\), and has its center on the line \(x+y=5\), follow these steps:

1. Let the center of the circle be \((h,k)\). Since the center lies on the line \(x+y=5\), we have:

\(h+k=5\) (Equation 1)

2. The general equation of a circle is \((x-h)^2+(y-k)^2=r^2\).
3. Since the circle passes through \((1,2)\), substitute these coordinates into the circle's equation:

\((1-h)^2+(2-k)^2=r^2\) (Equation 2)

4. Similarly, for the point \((4,3)\):

\((4-h)^2+(3-k)^2=r^2\) (Equation 3)

5. Subtract Equation 2 from Equation 3 to eliminate \(r^2\):

\(((4-h)^2+(3-k)^2) - ((1-h)^2+(2-k)^2) = 0\)

Expanding both sides and simplifying:

\[(4-h)^2 - (1-h)^2 + (3-k)^2 - (2-k)^2 = 0\]

\[(3)(4+1-2h)+(1)(3+2-2k)=0\]

\[15-6h+5-2k = 0\]

From here, simplify to get:

\[3h+k=10\] (Equation 4)

6. We now solve the system of equations (Equation 1 and 4):

Equation 1: \(h+k=5\)

Equation 4: \(3h+k=10\)

7. Subtract Equation 1 from Equation 4:

\[(3h+k)-(h+k)=10-5\]

\[2h=5\]

\[h=2.5\]

Substitute \(h=2.5\) into \(h+k=5\):

\[2.5+k=5\]

\[k=2.5\]

8. Substitute \(h=2.5\) and \(k=2.5\) into Equation 2 to find \(r^2\):

\((1-2.5)^2+(2-2.5)^2=r^2\)

\[(1.5)^2+(0.5)^2=r^2\]

\[2.25+0.25=r^2\]

\[r^2=2.5\]

9. The equation of the circle is thus:

\((x-2.5)^2+(y-2.5)^2=2.5\)

Therefore, the correct equation of the circle is \((x-2.5)^2+(y-2.5)^2=2.5\).

Answer 2

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: \[ (x - 2.5)^2 + (y - 2.5)^2 = 2.5 \]