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 determine the equation of the circle passing through \((1,2)\) and \((4,3)\) with its center on the line \(x+y=5\), execute the subsequent steps:

1. Designate the circle's center as \((h,k)\). Given that the center is on the line \(x+y=5\), the following equation is established:

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

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

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

4. Apply the same procedure for the point \((4,3)\):

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

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

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

Expand and simplify both sides:

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

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

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

Further simplification yields:

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

6. Solve the system comprising Equation 1 and Equation 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 ascertain \(r^2\):

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

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

\[2.25+0.25=r^2\]

\[r^2=2.5\]

9. Consequently, the equation of the circle is:

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

The definitive equation for the circle is \((x-2.5)^2+(y-2.5)^2=2.5\).