CBSE Class 10 Answered
Find the center of circle passing through the points (0,0) (-2,1) (-3,2). Find its radius.
Asked by | 10 Mar, 2013, 11:02: PM
Given : circle passing through the points (0,0) (-2,1) (-3,2)
To find : its centre and radius.
Equation of a circle: (x - h)2 + (y - k)2 = r2 where (h,k) is the center and r is the radius.
For the first problem, you need to find the values of h, k and r. Write 3 equationsreplacing x and y with the coordinates from each ordered pair (0,0) (-2,1) (-3,2). Then you will have a system of three equations containing the three variables you need to find.
(0 - h)2 + (0 - k)2 = r2...............................(a)
(-2 - h)2 + (1 - k)2= r2
(-3 - h)2 + (2 - k)2 = r2
Since all three expressions equal r2, you can set any two equal to each other.
=> (0 - h)2 + (0 - k)2 = (-2 - h)2 + (1 - k)2
For the first problem, you need to find the values of h, k and r. Write 3 equationsreplacing x and y with the coordinates from each ordered pair (0,0) (-2,1) (-3,2). Then you will have a system of three equations containing the three variables you need to find.
(0 - h)2 + (0 - k)2 = r2...............................(a)
(-2 - h)2 + (1 - k)2= r2
(-3 - h)2 + (2 - k)2 = r2
Since all three expressions equal r2, you can set any two equal to each other.
=> (0 - h)2 + (0 - k)2 = (-2 - h)2 + (1 - k)2
=> h2 +k2 = 4 + h2 + 4h + 1+k2 -2k
=> 4h-2k= -5........................(1)
similarily
=> (0 - h)2 + (0 - k)2 = (-3 - h)2 + (2 - k)2
=> h2+k2 = 9 + h2 +6h + 4 +k2 -4k
=>6h-4k = -13.........................................(2)
putting the value obtained of h from eq 1 to eq 2
=> 6((-5+2k)/4)-4k=-13
=> (-15/2) + 3k -4k = -13
=> k = 11/2.....................................(3)
putting the value of eq 3 in eq 2 , we get
=> 6h-4(11/2) = -13
=> 6h = 9
=> h= 3/2.....................(4)
from eq 3 and eq 4
the center of the circle required is (h,k ) i.e (3/2 , 11/2)
Putting the value of eq 3 and 4 in eq a , we get
=>(0 - (3/2))2 + (0 - (11/2))2 = r2
=> r2 = (9 /4) + (121/4)
=> r2 = 130/4
=> r = 11.4/2
=> r = 5.7 (approx) Answer
Answered by | 11 Mar, 2013, 12:37: AM
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