Additional Mathematics – Coordinate geometry of the circle | e-Consult

Answer 3

1. Let the centre be \((h,k)\) with the condition \(h+k=4\).
2. Since the circle passes through \((1,2)\) and \((5,6)\):

   • \((1-h)^{2}+(2-k)^{2}=r^{2}\)
   • \((5-h)^{2}+(6-k)^{2}=r^{2}\)

3. Subtract the two equations to eliminate \(r^{2}\):

   \((5-h)^{2}-(1-h)^{2}+(6-k)^{2}-(2-k)^{2}=0\)

4. Expand:

   \([25-10h+h^{2}]-[1-2h+h^{2}] + [36-12k+k^{2}] - [4-4k+k^{2}] =0\)

5. Simplify:

   \(24-8h + 32-8k =0\) → \(56-8(h+k)=0\)

6. Using \(h+k=4\): \(56-8(4)=56-32=24\neq0\).

   Re‑check algebra: actually after simplification we obtain

   \(8h+8k=56\) → \(h+k=7\).

   Combine with the given line \(h+k=4\) gives a contradiction, indicating a mis‑step.

7. Correct approach: use the perpendicular bisector of the segment joining the two points.

   Midpoint M = \(\bigl(\frac{1+5}{2},\frac{2+6}{2}\bigr) = (3,4)\).

   Slope of AB = \(\frac{6-2}{5-1}=1\); therefore the perpendicular bisector has slope \(-1\) and equation:

   \((y-4)= -1(x-3) \Rightarrow y = -x +7\).

8. Centre lies on both lines:

   \(\begin{cases}x+y=4\\ y=-x+7\end{cases}\) → substitute: \(x+(-x+7)=4\) → \(7=4\) (contradiction).

9. Hence the only possible centre satisfying both conditions is the intersection of the two lines, which does not exist; therefore the problem data is inconsistent.

Result: No circle can satisfy the given conditions.