Question 4550: i have to find the standard form of the equation of the specified circle and I'm told that the center is (-1,2) and the solution point is (0,0). can someone help me find the answer and show me how you got it?
Answer by ramanna(5)   (Show Source):
You can put this solution on YOUR website! Hi student,
you confused with the question..you might have told to find the equation of the circle whose center is (-1,2) and passes through the point (0,0).
the standard equation of the circle whose centre is (h,k) and radius is 'a'
can be writtin as
(x-h)^2 + (y-k)^2 = a^2 ( remember ^2 means square)
in the above problem centre is given which is (h,k)=(-1,2), that means h=-1 and k=2, but radius is not given... but it is told that the circle passing through (0,0) and its centre is (-1,2), the distance between these two points is the radius of the cirlce!! isnt it?? ..ya, that distance is the radius of the circle..
this distance(radius) can be found by distance formula..
that means.. root of( ((0-(-1))^2) + (0-2)^2)
= root of ( (0+1)^2) + (0-2)^2)
= root of ( 1^2 + ((-2)^2) )
= roof of (1 + 4)
= root of (5)
..this is equal to 'a' = radius..
therefore now we have, centre (h,k) = (-1,2) and radius = root of (5)
substituting this in the standard equation of the circle , we get..
(x-(-1))^2 + (y-2)^2 = (root of (5))^2
that means
(x+1)^2 + (y-2)^2 = 5
thats the answer..
its difficult to write here everything in mathematical language..try to understand it.. if not then write to me at ramanna11@rediffmail.com
|
|
|
