Question

Find the equation of the circle which circumscribes the triangle formed by the lines
(i) x + y + 3 = 0, x − y + 1 = 0 and x = 3
(ii) 2x + y − 3 = 0, x + y − 1 = 0 and 3x + 2y − 5 = 0
(iii) x + y = 2, 3x − 4y = 6 and x − y = 0.
(iv) y = x + 2, 3y = 4x and 2y = 3x.

Answer 1

In $∆$ABC:
(i) Let AB represent the line x + y + 3 = 0. ...(1)
Let BC represent the line x − y + 1 = 0. ...(2)
Let CA represent the line x = 3. ...(3)

Intersection point of (1) and (3) is $\left(3,-6\right)$.
Intersection point of (1) and (2) is (−2, −1).
Intersection point of (2) and (3) is (3, 4).

Therefore, the coordinates of A, B and C are $\left(3,-6\right)$, (−2, −1) and (3, 4), respectively.

Let the equation of the circumcircle be $x^2+y^2+2gx+2fy+c=0$.
It passes through A, B and C.

∴ $45+6g-12f+c=0$
$5-4g-2f+c=0$
$25+6g+8f+c=0$

$\therefore g=-3,f=1,c=-15$

Hence, the required equation of the circumcircle is $x^2+y^2-6x+2y-15=0$.


(ii) In $∆$ABC:
Let AB represent the line 2x + y − 3 = 0. ...(1)
Let BC represent the line x + y − 1 = 0. ...(2)
Let CA represent the line 3x + 2y − 5 = 0. ...(3)

Intersection point of (1) and (3) is (1, 1).
Intersection point of (1) and (2) is (2, −1).
Intersection point of (2) and (3) is (3, −2).

The coordinates of A, B and C are (1, 1), (2, −1) and (3, −2), respectively.

Let the equation of the circumcircle be $x^2+y^2+2gx+2fy+c=0$.
It passes through A, B and C.

∴ $2+2g+2f+c=0$
$5+4g-2f+c=0$
$13+6g-4f+c=0$

$\therefore g=\frac{-13}{2},f=\frac{-5}{2},c=16$

Hence, the required equation of the circumcircle is $x^2+y^2-13x-5y+16=0$.

(iii) In $∆$ABC:
Let AB represent the line x + y = 2. ...(1)
Let BC represent the line 3x − 4y = 6. ...(2)
Let CA represent the line x − y = 0. ...(3)

Intersection point of (1) and (3) is (1, 1).
Intersection point of (1) and (2) is (2, 0).
Intersection point of (2) and (3) is (−6, −6).

The coordinates of A, B and C are (1, 1), (2, 0) and (−6, −6), respectively.

Let the equation of the circumcircle be $x^2+y^2+2gx+2fy+c=0$.
It passes through A, B and C.
∴ $2+2g+2f+c=0$
$4+4g+c=0$
$72-12g-12f+c=0$

$\therefore g=2,f=3,c=-12$

Hence, the required equation of the circumcircle is $x^2+y^2+4x+6y-12=0$.

(iv)
In $∆$ABC:
(i) Let AB represent the line y = x + 2 ...(1)
Let BC represent the line 3y = 4x ...(2)
Let CA represent the line 2y = 3x ...(3)

Intersection point of (1) and (3) is (4, 6)
Intersection point of (1) and (2) is (6, 8).
Intersection point of (2) and (3) is (0, 0).

Therefore, the coordinates of A, B and C are (4, 6), (6, 8) and (0, 0) respectively.

Let the equation of the circumcircle be $x^2+y^2+2gx+2fy+c=0$.
It passes through A, B and C.

∴ $52+8g+12f+c=0$, $100+12g+16f+c=0$
and

$$\begin{gathered} 0++0+0+0+c=0 \\ \Rightarrow c=0 \end{gathered}$$

$\therefore g=-23,f=11,c=0$

Hence, the required equation of the circumcircle is $x^2+y^2-46x+22y=0$.