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# Find the center of circle passing through the points (0,0) (-2,1) (-3,2). Find its radius.

Asked by 10th March 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
=> 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 Expert 11th March 2013, 12:37 AM
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