# Given the circle center (1,1) and the point on the circle (3,5) what is the equation of the circle.

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### 2 Answers

The general equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.

Now we know that the center of the circle is (1, 1), point (3 , 5) lies on the circle.

Therefore the distance between the two is sqrt [( 3-1)^2 + (5-1)^2]

=> sqrt [ 4 + 16]

=> sqrt 20

So the equation if the circle is (x - 1)^2 + (y - 1)^2 = [sqrt 20]^2=> (x - 1)^2 + (y - 1)^2 = 20

**The required equation is (x - 1)^2 + (y - 1)^2 = 20.**

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 = 1 and k = 1.

We'll substitute them into the equation:

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

We'll determine the radius considering the constraint from enunciation that the circle is passing through the point (3,5).

If the circle is passing through the point (3,5), then the coordinates of the point are verifying the equation of the circle:

(3 - 1)^2 + (5 - 1)^2 = r^2

2^2 + 4^2 = r^2

4 + 16 = r^2

20 = r^2

r = 2sqrt5

Note: we'll reject th negative value of radius -2sqrt5.

**The equation of the circle is:**

**(x - 1)^2 + (y - 1)^2 = 20**