Find the center and radius of the circle whose equation is given by: x^2 + y^2 - 10x - 2y + 22 = 0
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calendarEducator since 2008
write3,662 answers
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x62 + y62 - 10x - 2y + 22 = 0
First we need to rewrtie the equation in the standard formula:
(x-a)^2 + (y-b)^2 = r^2 such that:
(a, b) is the center and r is the radius.
To rewrite we will need to complete the squares:
x^2 -10x +25 -25 + y^2 - 2x + 1 -1 + 22 = 0
( x-5)^2 + (y-2)^2 - 25 - 1 + 22 = 0
==> (x-5)^2 + (y-2)^2 - 4 = 0
==> (x-5)^2 + (y-2)^2 = 4
Then we conclude that the center is:
(5, 2) and the radius r = 2
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We have to find center and radius of the circle with the equation x^2 + y^2 - 10x - 2y + 22 = 0.
Now the standard equation of a circle with center (a,b) and radius r is : (x-a)^2 + (y-b)^2 = r^2.
We write x^2 + y^2 - 10x - 2y + 22 = 0
=> x^2 - 10x + y^2 - 2y + 22 =0
Now complete the squares, (keep in mind that (a-b)^2 = a^2+ b^2 - 2ab)
=> x^2 - 10x + 25 + y^2 - 2y + 1 + 22 -25 -1 =0
=> (x - 5)^2 + (y - 2)^2 -4 = 0
=> (x - 5)^2 + (y - 2)^2 = 4
=> (x - 5)^2 + (y - 2)^2 = 2^2
Therefore we get the center as (5, 2) and the radius as 2.
To find the centre and radius of the circle given by: x^2 + y^2 - 10x - 2y + 22 = 0.
solution:
If (h,k) are the centre of a circle and r is the radius, then its equation is given by:
(x-h)^2+(y-k)^2 = r^2...(1).
So the given equation x^2 + y^2 - 10x - 2y + 22 = 0 could therefore be rearranged as:
(x^2-10x +5^2)+(y^2-2y+2^2) +22- (5^2+2^2) = 0.
(x-5)^2+(y-1)^2 +22-27 = 0.
(x-5)^2+(y-1)^2 = 5 = (sqrt(5))^2.
(x-5)^2+(y-1)^2 = (sqrt5)^2..(2).
Identifying the equations (1) and (2) , we notice that :
(h,k) = (5,1) and r = sqrt5.
Therefore the centre of the circle x^2 + y^2 - 10x - 2y + 22 = 0 is (5,1) and radius is sqrt5.
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