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) $2 x + y - 3 = 0, x + y - 1 = 0$ and $3 x + 2 y - 5 = 0$
(iii) $x + y = 2, 3 x - 4 y = 6$ and $x - y = 0.$
(iv) $y = x + 2, 3 y = 4 x$ and $2 y = 3 x$ .

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Solution

The given equation of lines
(i) $x + y = - 3 \dots (i) x - y = \dots (i i) x = 3 \dots (i i i)$
Let A, B, C are the point of intersection of lines (i) and (ii), (ii) and (iii) and (i) respectively
$∴ A = (- 2, - l), B (3, 4)$ and $C = (3, - 6)$
Now A circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \dots (A)$
circumscribing the $Δ A B C$

$∴ 4 + 1 - 4 g - 2 f + c = 0 \dots (i v) 9 + 16 + 6 g + 8 f + c = 0 \dots (v) 9 + 36 + 6 g - 12 f + c = 0 \dots (v i)$
Solving (iv), (v) and (vi) we get,
g =-3,f = 1, and c = -15
from (A),
The equation of required circle is
$x^{2} + y^{2} - 6 x + 2 y - 15 = 0$

The equation of lines are
$2 x + y - 3 = 0 x + y - 1 = 0 3 x + 2 y - 5 = 0$
The equation of lines can be written as:
$2 x + y = 3 \dots (i) x + y = 1 \dots (i i) 3 x + 2 y = 5 \dots (i i i)$
Let A, B and C arc the points of intersection of the lines (i) and (ii),

(ii) and (iii) and (iii) and (i) respectively.
$∴ A = (2, - 1) B = (3, - 2) C = (1, 1)$
Let $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
be the circle that circumscribing $Δ A B C$ .
$∴ 4 + 1 + 4 g - 2 f + c = 0 \dots (i v) 9 + 4 + 6 g - 4 f + c = 0 \dots (v) 1 + 1 + 2 g + 2 f + c = 0 \dots (v i)$
Solving (iv), (v) and (vi) we get,
$g = - \frac{13}{2}, f = - \frac{5}{2} c = 16$
from (A), The required circle is
$x^{2} + y^{2} - 13 x - 5 y + 16 = 0$

(iii) The given equation of lines are
$x + y = 2 \dots (i) 3 x - 4 y = 6 \dots (i i) x - y = 0 \dots (i i i)$
Let A, B and C arc the points of intersection of the lines (i) and (ii), (ii) and (iii) and (iii) and (i) respectively.
$∴ A = (2, 0) B = (6, - 6) C = (1, 1)$
Let $x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \dots (A)$
be the circle that circumscribing $Δ A B C$ .
$∴ 4 + 4 g + c = 0 \dots (i v) 36 + 36 - 12 g - 12 f + c = 0 \dots (v) 1 + I + 2 g + 2 f + c = 0 \dots (v i)$
Solving (iv), (v) and (vi) we get,\\
g = 2, f = 3 and c = -12
from (A),
The required circle is
$x^{2} + y^{2} + 4 x + 6 y - 12 = 0$

(iv) Given equation of line are
$y = x + 2 \dots (i) 3 y = 4 x \dots (i i) 2 y = 3 x \dots (i i i)$
From Eqs. (i) and (ii),
$\frac{4 x}{3} = x + 2 \Rightarrow 4 x = 3 x + 6 + 6 \Rightarrow x = 6$
On putting x=6 in Eq. (i), we get
$y = 8$
$∴$ Point, A = (6, 8)
From Eqs. (i) and (iii),
$\frac{3 x}{2} = x + 2 \Rightarrow 3 x = 2 x + 4 \Rightarrow x = 4$
When, x= 4, then y = 6
$∴$ Point, B = (4, 6)
From Eqs. (ii) and (iii)
$x_{1} = 0, y = 0$
Now, $C = 0 (0, 0)$

Let the equation of circle is
$x^{2} + y^{2} + 2 g x n + 2 f y + c = 0$
Since, the points A (6, 8), B (4, 6) and C (0, 0) lie on this circle. $36 + 64 + 12 g + 16 f + c = 0$
$\Rightarrow 12 g + 16 f + c = - 100 \dots (i v)$
and $16 + 36 + 8 g + 12 f + c = 0$
$\Rightarrow 8 g + 12 f + c = - 52 \dots (v) \Rightarrow c = 0 \dots (v i)$
From Eqs. (iv), (v) and (vi),
$12 g + 16 f = - 100 \Rightarrow 3 g + 4 f + 25 = 0 \Rightarrow 2 g + 3 f + 13 = 0 \Rightarrow \frac{g}{52 - 75} = \frac{f}{50 - 39} = \frac{1}{9 - 8} \Rightarrow \frac{g}{- 23} = \frac{f}{11} = \frac{1}{1} \Rightarrow g = - 23, f = 11$
So, the equation of circle is
$x^{2} + y^{2} - 46 x + 22 y + 0 = 0 \Rightarrow x^{2} + y^{2} - 46 x + 22 y = 0$

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