Question

Find the equation of the circle passing through the points \((4, 1) \)and \((6, 5)\) and whose center is on line \(4x + y = 16.\)

Answer 1

Let the equation of the required circle be \((x - h)^2 + (y - k)^2 = r^2 .\)
Since the circle passes through points \((4, 1)\) and \((6, 5),\)

\((4 - h)^2 + (1 - k)^ 2 = r^2 … (1)\)
\((6 - h) ^2 + (5 - k) ^2 = r^2 … (2)\)

Since the center \((h, k)\) of the circle lies on the line 
\(4x + y = 16,  \)

\(4h + k = 16 … (3)\)
From equations (1) and (2), we obtain 
\((4 - h)^2 + (1 - k)^2 = (6 - h)^2 + (5 - k)^2 \)
\(⇒ 16 - 8h + h^2 + 1 - 2k + k^2 = 36 - 12h + h^2 + 25 - 10k + k^2 \)
\(⇒ 16 - 8h + 1 - 2k = 36 - 12h + 25 - 10k \)
\(⇒ 4h + 8k = 44 \)
⇒ h + 2k = 11 … (4)\(\)

On solving equations (3) and (4), we obtain \(h = 3 \)and k = 4. \(\)
On substituting the values of h and k in equation (1), we obtain 
\((4 - 3)^2 + (1 - 4)^2 = r^2 \)
\(⇒ (1)^2 + (- 3) ^2 = r ^2 \)
\(⇒ 1 + 9 = r^2 \)
\(⇒ r ^2 = 10\)
\(⇒ r = √10\)

Thus, the equation of the required circle is
\((x - 3) ^2 + (y - 4) ^2 = (√10)^2 \)
\(x^2 - 6x + 9 + y ^2 - 8y + 16 = 10 \)
\(x^2 + y^2 - 6x - 8y + 15 = 0\)\(\)
∴ The equation of the required circle is \(x^2 + y^2 – 6x – 8y + 15 = 0\) (Ans)