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)