Question

Find the equations of circles which passes through the vertices of the triangle formed by the line
x + y + 1 = 0 , 3x +y - 5 = 0 and 2x + y - 5 = 0 z

Answer 1

See (j) three methods P. 851.
Solving the sides of the triangle in pairs the required vertices are (3, -4), (0, 5) and (6, -7). Now proceeding as in Q. 19 P. 856-870 the circle is
$x^2+y^2$ - 30x 10 y + 25 = 0.
Alternative : Without finding the points of intersection, let the equation of the circle through the points of intersection of given lines be
$l_1{l}_2+\lambda \left(l_2{l}_3\right)+\mu \left(l_3{l}_1\right)=0$
Apply the condition that the above equation represents a circle i.e. coeff. of $x^2$ = coeff. of $y^2$ and coeff. of xy = 0. This will the two equations in $\lambda$ and $\mu$. Solve and put in (1). All this is explained with a numerical example given in part (b) below.