Question

Prove that the equation of the circle circumscribing the triangle formed by the sides whose equations are 2x + y - 3 = 0 , 3x - y - 7 = 0 and x - 2y + 1 = 0 is $x^2+y^2-$ 5x - y + 4 = 0

Answer 1

Let the equation of the lines be P = 0, Q = 0, R = 0. Consider the equation QR + $\lambda$RP + PQ = 0
(3x- y - 7) (x- 2y + 1) + $\lambda$(x - 2y + 1) (2x + y - 3)
$\mu$ (2x+ y - 3) (3x - y - 7) = 0 (1)
The above equation is satisfied by Q = 0, R = 0 i e point A and similarly by points B and C Hence the curve whatsoever it may be passes through the points A B and C Since it is to be a circle.
$\therefore$ Coeff. of $x^2$ = Coeff of $y^2$ and Coeff. of xy = 0 in (1)
$\therefore 3+2\lambda +6\mu =2-2\lambda -\mu$
and -7 + $\lambda$ (- 3) + $\mu$ (1) = 0
or 4$\lambda$ + 7 $\mu$ = -1 and 3 $\lambda$ - $\mu$ = -7
Solving the above, we get $\lambda$ = -2, $\mu$ = 1.
Putting the values of $\lambda$ and $\mu$ in (1), we get the result as given.