Find the equation of the circle passing through the points (4, 1)
Question:

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

Solution:

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),

$(4-h)^{2}+(1-k)^{2}=r^{2} .$

$(6-h)^{2}+(5-k)^{2}=r^{2} .$

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

$4 h+k=16$   (3)

From equations (1) and (2), we obtain

$(4-h)^{2}+(1-k)^{2}=(6-h)^{2}+(5-k)^{2}$

$\Rightarrow 16-8 h+h^{2}+1-2 k+k^{2}=36-12 h+h^{2}+25-10 k+k^{2}$

$\Rightarrow 16-8 h+1-2 k=36-12 h+25-10 k$

$\Rightarrow 4 h+8 k=44$

$\Rightarrow h+2 k=11 \ldots(4)$

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

$(4-3)^{2}+(1-4)^{2}=r^{2}$

$\Rightarrow(1)^{2}+(-3)^{2}=r^{2}$

$\Rightarrow 1+9=r^{2}$

$\Rightarrow r^{2}=10$

$\Rightarrow r=\sqrt{10}$

Thus, the equation of the required circle is

$(x-3)^{2}+(y-4)^{2}=(\sqrt{10})^{2}$

$x^{2}-6 x+9+y^{2}-8 y+16=10$

$x^{2}+y^{2}-6 x-8 y+15=0$