# Find the center and radius of the circle whose equation is given by: (5 - x)^2 + (y - 1)^2 = 4

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### 3 Answers

(5-x)^2 + (y-1)^2 = 4

We know that the standrad form for the circle equation is given by the formula:

( x- a)^2 + (y- b)^2 = r^2

such that:

(a, b) is a point such that the point is the center of the circle.

r is the length of the radius of the circle .

Then we will rewrtie the equation:

We can wrtie 4 = 2^2

( x- 5)^2 + (y-1)^2 = 2^2

Now comparing both equation we conclude that:

**The center of the circle is ( 5, 1)**

**And the radius is 2 units**.

We know that the equation of the circle whose centre is (h, k) and radius r is given by:

(x-h)^2+(y-k)^2 = r^2....(1).

The given equation (5-x)^2 +(y-1)^2 = 4 could be rewitten as:

(x-5)^2 +(y-1)^2 = 2^2. ....(2), as (5-x)^2 = (x-5)^2 .

Now we can identify equations (1) and (2) and equate the like terms:

(x-h)^2 = (x-5)^2 gives h= 5.

(y-k)^2 = (y-1)^2 gives k = 1.

r^2 = 2^2 gives r = 2.

Therefore the centre of the given circle is (5 , 1) and the radius is 2.

The equation of the circle, whose radius is r and center of the circle is C(a,b):

(x-m)^2 + (y-n)^2 = r^2

Now, we'll write the given equation:

(5 - x)^2 + (y - 1)^2 = 4

[-(x-5)]^2 + (y - 1)^2 = 2^2

We'll identify the coordinates of the center of the circle as:

x coordinate:m = 5 and

y coordinate: n = 1

**The center of the circle is C(5,1) and the radius is r = 2.**