Find the equation of the circle passing through the intersection points of three given equations :

3x + y = -13

L1 : 3x + y + 13 = 0

L2 : x + 2y – 4 = 0

L3 : x - y - 1 = 0

Rewrite the equations in General Form.

Let L1 and L2 meet at P,

L2 and L3 meet at Q

and L3 and L1 meet at R.

Form the system of conic:

L1 L2 + a L2 L3 + b L3 L1 = 0

This system of equation must pass through P, Q, R.

(3x+y+13)(x+2y-4)+a(x+2y-4)(x-y-1)+b(x-y-1)(3x+y+13)=0

+(22 + 2a – 14b) y –52 + 4a – 13b = 0 ….…(*)

It is interesting that we do not find the intersection points P, Q, R.

This is an equation of degree 2 and is a general conic which passes through P, Q and R. Why?

(*) is a circle, therefore:

(1) Coefficient of x-square term = coeff. of y-square term

(2) Coefficient of xy-term = 0

that is,

(1) 3 + a + 3b = 2 - 2a – b ….. (1)

(2) 7 + a –2b = 0 ……………...(2)

From (2), a = 2b – 7…………. (3)

Subst. (3) in (1), \ b = 2 ……… (4)

Subst. (4) in (3), \ a = -3 ……... (5)

These are the conditions for a general conic to be a circle.

A simultaneous equation with a, b is got.

Subst. (4), (5) in (*), the required equation is:

Bingo! We find the circle.