# 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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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.

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.**

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**