Q.Find the equation of the circle passing through the points $(6,5)$ and $(-1,4)$, given that its centre lies on the line $x-y=2$.

Let the circle be $x^2+y^2+2gx+2fy+c=0$.

Substituting $(6,5)$: $36+25+12g+10f+c=0\Rightarrow 12g+10f+c=-61.\quad(i)$

Substituting $(-1,4)$: $1+16-2g+8f+c=0\Rightarrow -2g+8f+c=-17.\quad(ii)$

The centre $(-g,-f)$ lies on $x-y=2$: $(-g)-(-f)=2\Rightarrow f=g+2.\quad(iii)$

Subtracting (ii) from (i): $14g+2f=-44\Rightarrow 7g+f=-22.\quad(iv)$

Substituting (iii) into (iv): $7g+(g+2)=-22\Rightarrow 8g=-24\Rightarrow g=-3$, so $f=-1$.

From (ii): $-2(-3)+8(-1)+c=-17\Rightarrow 6-8+c=-17\Rightarrow c=-15.$

So the equation is $x^2+y^2-6x-2y-15=0$, i.e. centre $(3,1)$, radius $\sqrt{9+1+15}=\sqrt{25}=5$.

Substitute both given points into the general circle equation to get two linear equations in $g,f,c$; use the centre-on-a-line condition as a third equation; solve the resulting system simultaneously.

Substituting the line condition on $(h,k)$ directly (as $h-k=2$) but then forgetting $h=-g,\ k=-f$, so writing $g-f=2$ instead of the correct $-g+f=2$; arithmetic slips solving the 3-equation linear system.