# Given the point (4,1) on the circle that have the center (2,1), write the equation of circle?

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### 2 Answers

The equation of a circle requires the center and the radius of the circle. We have the center as (2 ,1). For the radius we use the fact that the point (4, 1) is on the circle.

The distance between (4 , 1) nad (2,1) is given by

sqrt [( 1-1)^2 + (4-2)^2] = sqrt 4 = 2

We can now write the equation of the circle in the form (x - 2)^2 + (y - 1)^2 = 4

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

**Therefore the equation of the circle is x^2 + y^2 - 4x - 2y +1 =0**

We'll write the equation, containing perfect squares, 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 determine the radius, considering the condition from enunciation,namely that the the point (4,1) is located on the given circle.

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, whose center is (2,1), is:**

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