Question -
Answer -
(i) (5, 7), (8, 1) and (1, 3)
By using the standardform of the equation of the circle:
x2┬а+ y2┬а+2ax + 2by + c = 0 тАж.. (1)
Firstly let us findthe values of a, b and c
Substitute the givenpoint (5, 7) in equation (1), we get
52┬а+ 72┬а+2a (5) + 2b (7) + c = 0
25 + 49 + 10a + 14b +c = 0
10a + 14b + c + 74 =0тАж.. (2)
By substituting thegiven point (8, 1) in equation (1), we get
82┬а+ 12┬а+2a (8) + 2b (1) + c = 0
64 + 1 + 16a + 2b + c= 0
16a + 2b + c + 65 =0тАж.. (3)
Substituting the point(1, 3) in equation (1), we get
12┬а+ 32┬а+2a (1) + 2b (3) + c = 0
1 + 9 + 2a + 6b + c =0
2a + 6b + c + 10 =0тАж.. (4)
Now by simplifying theequations (2), (3), (4) we get the values
a = -29/6, b = -19/6,c = 56/3
Substituting thevalues of a, b, c in equation (1), we get
x2┬а+ y2┬а+2 (-29/6)x + 2 (-19/6) + 56/3 = 0
x2┬а+ y2┬атАУ29x/3 тАУ 19y/3 + 56/3 = 0
3x2┬а+3y2┬атАУ 29x тАУ 19y + 56 = 0
тИ┤┬аThe equation ofthe circle is 3x2┬а+ 3y2┬атАУ 29x тАУ 19y + 56 = 0
(ii)┬а(1, 2), (3, тАУ 4) and(5, тАУ 6)
By using the standardform of the equation of the circle:
x2┬а+ y2┬а+2ax + 2by + c = 0 тАж.. (1)
Substitute the points(1, 2) in equation (1), we get
12┬а+ 22┬а+2a (1) + 2b (2) + c = 0
1 + 4 + 2a + 4b + c =0
2a + 4b + c + 5 = 0тАж..(2)
Substitute the points(3, -4) in equation (1), we get
32┬а+(- 4)2┬а+ 2a (3) + 2b (- 4) + c = 0
9 + 16 + 6a тАУ 8b + c =0
6a тАУ 8b + c + 25 =0тАж.. (3)
Substitute the points(5, -6) in equation (1), we get
52┬а+(- 6)2┬а+ 2a (5) + 2b (- 6) + c = 0
25 + 36 + 10a тАУ 12b +c = 0
10a тАУ 12b + c + 61 =0тАж.. (4)
Now by simplifying theequations (2), (3), (4) we get
a = тАУ 11, b = тАУ 2, c =25
Substitute the valuesof a, b and c in equation (1), we get
x2┬а+ y2┬а+2(- 11)x + 2(- 2) + 25 = 0
x2┬а+ y2┬атАУ22x тАУ 4y + 25 = 0
тИ┤┬аThe equation ofthe circle is x2┬а+ y2┬атАУ 22x тАУ 4y + 25 = 0
(iii) (5, -8), (-2, 9) and (2, 1)
By using the standardform of the equation of the circle:
x2┬а+ y2┬а+2ax + 2by + c = 0 тАж.. (1)
Substitute the point(5, -8) in equation (1), we get
52┬а+(- 8)2┬а+ 2a(5) + 2b(- 8) + c = 0
25 + 64 + 10a тАУ 16b +c = 0
10a тАУ 16b + c + 89 =0тАж.. (2)
Substitute the points(-2, 9) in equation (1), we get
(- 2)2┬а+92┬а+ 2a(- 2) + 2b(9) + c = 0
4 + 81 тАУ 4a + 18b + c= 0
-4a + 18b + c + 85 =0тАж.. (3)
Substitute the points(2, 1) in equation (1), we get
22┬а+ 12┬а+2a(2) + 2b(1) + c = 0
4 + 1 + 4a + 2b + c =0
4a + 2b + c + 5 = 0тАж..(4)
By simplifyingequations (2), (3), (4) we get
a = 58, b = 24, c = тАУ285.
Now, by substitutingthe values of a, b, c in equation (1), we get
x2┬а+ y2┬а+2(58)x + 2(24) тАУ 285 = 0
x2┬а+ y2┬а+116x + 48y тАУ 285 = 0
тИ┤┬аThe equation ofthe circle is x2┬а+ y2┬а+ 116x + 48y тАУ 285 = 0
(iv) (0, 0), (-2, 1) and (-3, 2)
By using the standardform of the equation of the circle:
x2┬а+ y2┬а+2ax + 2by + c = 0 тАж.. (1)
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.. (2)
Substitute the points(-2, 1) in equation (1), we get
(- 2)2┬а+12┬а+ 2a(- 2) + 2b(1) + c = 0
4 + 1 тАУ 4a + 2b + c =0
-4a + 2b + c + 5 =0тАж.. (3)
Substitute the points(-3, 2) in equation (1), we get
(- 3)2┬а+22┬а+ 2a(- 3) + 2b(2) + c = 0
9 + 4 тАУ 6a + 4b + c =0
-6a + 4b + c + 13 =0тАж.. (4)
By simplifying theequations (2), (3), (4) we get
a = -3/2, b = -11/2, c= 0
Now, by substitutingthe values of a, b, c in equation (1), we get
x2┬а+ y2┬а+2(-3/2)x + 2(-11/2)y + 0 = 0
x2┬а+ y2┬атАУ3x тАУ 11y = 0
тИ┤┬аThe equation ofthe circle is x2┬а+ y2┬атАУ 3x тАУ 11y = 0