# find the equation of the circle with the center (2,1) if the circle passes through (4,1)

*print*Print*list*Cite

### 2 Answers

The equation of a circle with centre and radius r is (h,k) is (x-h)^2+(y-k)^2 = r^2.

Therefore the equation of the circle whose centre is (2,1) is (x-2)^2 + (y-1)^2 = r^2......(1). Since this circle passes through (4,1), the point (4,1) should satisfy the equation of the circle (1): So,

(4-2)^2 +(1-1)^2 = r^2

2^2+0 = r^2.

Therefore r^2 = 4.

So we rreplace r^2 = 4 in (1):

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

x^2-4x+4+y^2-2y+1 = 4.

x^2+y^2-4x-2y+5 = 4.

We rearrange write the equation of the circle in the standard form:

x^2+y^2-4x-2y+1 = 0.

We'll write the equation of the circle:

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

The center of the circle has the coordinates C(h ; k).

We know, from enunciation, that h = 2 and k = 1.

We'll substitute them into the equation:

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

We'll find the radius considering the condition from enunciation,namely that the circle is passing through the point (4,1).

If the circle is passing through the point (4,1), then the coordinates of the point are verifying the equation of the circle:

(4 - 2)^2 + (1 - 1)^2 = r^2

2^2 + 0^2 = r^2

r = 2

**The equation of the circle is:**

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