Find the equation of the circle which passes through the points (4, 1),(6, 5) and whose centre lies on 4x + 3y – 24 = 0

let equation of the required circle be
${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$
(4,1) lies on it
$\begin{array}{l}\Rightarrow 4^2+1^2+2\mathrm{g}\left(4\right)+2\mathrm{f}\left(1\right)+\mathrm{c}=0\\ \Rightarrow 8\mathrm{g}+2\mathrm{f}+\mathrm{c}=-17\dots ..\left(1\right)\end{array}$
(6,5) lies on it
$\begin{array}{l}\Rightarrow 6^2+5^2+2\mathrm{g}\left(6\right)+2\mathrm{f}\left(5\right)+\mathrm{c}=0\\ \Rightarrow 36+25+12\mathrm{g}+10\mathrm{f}+\mathrm{c}=0\\ \Rightarrow 12\mathrm{g}+10\mathrm{f}+\mathrm{c}=-61\dots .\left(2\right)\end{array}$
Centre (–g,–f) lies on the line 4x + 3y - 24 = 0
$\begin{array}{l}\Rightarrow 4(-\mathrm{g})+3(-\mathrm{f})-24=0\\ \Rightarrow -4\mathrm{g}-3\mathrm{f}-24=0\Rightarrow 4\mathrm{g}+3\mathrm{f}+24=0\end{array}$
Solving (1) & (2) (1)−(2)⇒−4g−8f =44→(4)
Solving (4) & (3)
(3)+ (4)⇒$-5\mathrm{f}=20\Rightarrow \mathrm{f}\equiv -4$
Put in (3) $4\mathrm{g}+3(-4)+24=0$
$\begin{array}{l}\Rightarrow 4\mathrm{g}-12+24=0\\ \Rightarrow 4\mathrm{g}=-12\Rightarrow \mathrm{g}=-3\end{array}$
Put f,g in $\text{ (1) }8(-3)+2(-4)+\mathrm{c}=-17$
$\begin{array}{l}\Rightarrow -24-8+\mathrm{c}=-17\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow -32+\mathrm{c}=-17\\ \Rightarrow \mathrm{c}=-17+32\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow \mathrm{c}=15\end{array}$
$\therefore$Equation of the required circle is
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2+2(-3)\mathrm{x}+2(-4)\mathrm{y}+15=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-8\mathrm{y}+15=0\end{array}$