Question -
Answer -
(i) x + y + 3 = 0, x тАУ y + 1 = 0 and x = 3
Given:
The lines x + y + 3 =0
x тАУ y + 1 = 0
x = 3
On solving these lineswe get the intersection points A (-2, -1), B (3, 4), C (3, -6)
So by using thestandard form of the equation of the circle:
x2┬а+ y2┬а+2ax + 2by + c = 0тАж.. (1)
Substitute the points(-2, -1) in equation (1), we get
(- 2)2┬а+(- 1)2┬а+ 2a(-2) + 2b(-1) + c = 0
4 + 1 тАУ 4a тАУ 2b + c =0
5 тАУ 4a тАУ 2b + c = 0
4a + 2b тАУ c тАУ 5 = 0тАж..(2)
Substitute the points(3, 4) in equation (1), we get
32┬а+ 42┬а+2a(3) + 2b(4) + c = 0
9 + 16 + 6a + 8b + c =0
6a + 8b + c + 25 =0тАж.. (3)
Substitute the points(3, -6) in equation (1), we get
32┬а+(- 6)2┬а+ 2a(3) + 2b(- 6) + c = 0
9 + 36 + 6a тАУ 12b + c= 0
6a тАУ 12b + c + 45 =0тАж.. (4)
Upon simplifyingequations (2), (3), (4) we get
a = тАУ 3, b = 1, c =-15.
Now by substitutingthe values of a, b, c in equation (1), we get
x2┬а+ y2┬а+2(- 3)x + 2(1)y тАУ 15 = 0
x2┬а+ y2┬атАУ6x + 2y тАУ 15 = 0
тИ┤┬аThe equation ofthe circle is x2┬а+ y2┬атАУ 6x + 2y тАУ 15 = 0.
(ii) 2x + y тАУ 3 = 0, x + y тАУ 1 = 0 and 3x + 2y тАУ 5 = 0
Given:
The lines 2x + y тАУ 3 =0
x + y тАУ 1 = 0
3x + 2y тАУ 5 = 0
On solving these lineswe get the intersection points A(2, тАУ 1), B(3, тАУ 2), C(1,1)
So by using thestandard form of the equation of the circle:
x2┬а+ y2┬а+2ax + 2by + c = 0тАж.. (1)
Substitute the points(2, -1) in equation (1), we get
22┬а+(- 1)2┬а+ 2a(2) + 2b(- 1) + c = 0
4 + 1 + 4a тАУ 2b + c =0
4a тАУ 2b + c + 5 = 0тАж..(2)
Substitute the points(3, -2) in equation (1), we get
32┬а+(- 2)2┬а+ 2a(3) + 2b(- 2) + c = 0
9 + 4 + 6a тАУ 4b + c =0
6a тАУ 4b + c + 13 =0тАж.. (3)
Substitute the points(1, 1) in equation (1), we get
12┬а+ 12┬а+2a(1) + 2b(1) + c = 0
1 + 1 + 2a + 2b + c =0
2a + 2b + c + 2 = 0тАж..(4)
Upon simplifyingequations (2), (3), (4) we get
a = -13/2, b = -5/2, c= 16
Now by substitutingthe values of a, b, c in equation (1), we get
x2┬а+ y2┬а+2 (-13/2)x + 2 (-5/2)y + 16 = 0
x2┬а+ y2┬атАУ13x тАУ 5y + 16 = 0
тИ┤┬аThe equation ofthe circle is x2┬а+ y2┬атАУ 13x тАУ 5y + 16 = 0
(iii) x + y = 2, 3x тАУ 4y = 6 and x тАУ y = 0
Given:
The lines x + y = 2
3x тАУ 4y = 6
x тАУ y = 0
On solving these lineswe get the intersection points A(2,0), B(- 6, тАУ 6), C(1,1)
So by using thestandard form of the equation of the circle:
x2┬а+ y2┬а+2ax + 2by + c = 0тАж.. (1)
Substitute the points(2, 0) in equation (1), we get
22┬а+ 02┬а+2a(2) + 2b(0) + c = 0
4 + 4a + c = 0
4a + c + 4 = 0тАж.. (2)
Substitute the point(-6, -6) in equation (1), we get
(- 6)2┬а+(- 6)2┬а+ 2a(- 6) + 2b(- 6) + c = 0
36 + 36 тАУ 12a тАУ 12b +c = 0
12a + 12b тАУ c тАУ 72 =0тАж.. (3)
Substitute the points(1, 1) in equation (1), we get
12┬а+ 12┬а+2a(1) + 2b(1) + c = 0
1 + 1 + 2a + 2b + c =0
2a + 2b + c + 2 = 0тАж..(4)
Upon simplifyingequations (2), (3), (4) we get
a = 2, b = 3, c = тАУ12.
Substituting thevalues of a, b, c in equation (1), we get
x2┬а+ y2┬а+2(2)x + 2(3)y тАУ 12 = 0
x2┬а+ y2┬а+4x + 6y тАУ 12 = 0
тИ┤┬аThe equation ofthe circle is x2┬а+ y2┬а+ 4x + 6y тАУ 12 = 0
(iv) y = x + 2, 3y = 4x and 2y = 3x
Given:
The lines y = x + 2
3y = 4x
2y = 3x
On solving these lineswe get the intersection points A(6,8), B(0,0), C(4,6)
So by using thestandard form of the equation of the circle:
x2┬а+ y2┬а+2ax + 2by + c = 0тАж.. (1)
Substitute the points(6, 8) in equation (1), we get
62┬а+ 82┬а+2a(6) + 2b(8) + c = 0
36 + 64 + 12a + 16b +c = 0
12a + 16b + c + 100 =0тАжтАж (2)
Substitute the points(0, 0) in equation (1), we get
02┬а+ 02┬а+2a(0) + 2b(0) + c = 0
0 + 0 + 0a + 0b + c =0
c = 0тАж.. (3)
Substitute the points(4, 6) in equation (1), we get
42┬а+ 62┬а+2a(4) + 2b(6) + c = 0
16 + 36 + 8a + 12b + c= 0
8a + 12b + c + 52 =0тАж.. (4)
Upon simplifyingequations (2), (3), (4) we get
a = тАУ 23, b = 11, c =0
Now by substitutingthe values of a, b, c in equation (1), we get
x2┬а+ y2┬а+2(- 23)x + 2(11)y + 0 = 0
x2┬а+ y2┬атАУ46x + 22y = 0
тИ┤┬аThe equation ofthe circle is x2┬а+ y2┬атАУ 46x + 22y = 0