Question

Find the circumcentre of the triangle whose side are
$3x-y-5=0,x+2y-4=0,5x+3y+1=0$

Answer 1

Find the circumcentre of $△$

$3x-y-5=\mathrm{0....}\left(i\right)$

$x+2y-4=\mathrm{0.....}\left(ii\right)$

$5x+3y+1=\mathrm{0.....}\left(iii\right)$

First find the vertices of the $△$

Solving (i) and (ii) simultaneously

$x_1=2,y_1=1$ $\therefore A(2,1)$

Similarly solving (ii) and (iii) simultaneously

$x_2=-2,y_2=3$

$\therefore$ $B(-2,3)$

Finally from (i) and (iii) $x_3=1,y_3=-2$ $C(1,-2)$

Let the equation of circle through $A,B,C$ be $x^2+y^2+2gx+2fy+c=0$

put the coordinates of $A(2,1)$

$4+1+4g+2g+c=0$

$4g+2f+c=-\mathrm{5....}\left(iv\right)$

Similarly by putting coordinates of $B(-2,3)$ in the circle equation we get

$4g-6g-c=\mathrm{13....}\left(v\right)$

and putting $C(1,-2)$

$2g-4f+c=-\mathrm{5....}\left(vi\right)$

Solving (iv), (v), (vi) simultaneously

$f=-\frac{2}{7},g=\frac{6}{7},c=-\frac{5.5}{7}$

The circumcentre of the circle $\equiv (-g,-f)$

$\equiv (-\frac{6}{7},\frac{2}{7})$