# Write the circle equation if it passes through the point (5,4) and the center of circle is (2,0).

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### 2 Answers

The equationof a circle with centre (h,k) and radius r is given by:

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

Therefore the circle with centre (2,0) is given by:

(x-2)^2 +(y-0)^2 = r^2.....(1).

So the circle at (1) passes through (5,4). So the point (5,4) should satisfy the equation at (2):

(5-2)^2+ (4)^2 = r^2 . So this gives: r^2 = 9+16 = 25. So substitute r = 25 in eq (1) and we rewite the equation:

(x-2)^2+y^2 = 25. We convert into standard form of equation as below by rearraging:

x^2-4x +4 +y^2-25 = 0.

x^2+y^2-4x-25 = 0 which is the required equation with centre (2,0) and passing through (5,4).

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 = 0.

We'll substitute them into the equation:

(x - 2)^2 + (y - 0)^2 = r^2

We'll determine the radius considering the constraint from enunciation that the circle is passing through the point (5,4).

If the circle is passing through the point (5,4), then the coordinates of the point are verifying the equation of the circle:

(5 - 2)^2 + (4 - 0)^2 = r^2

3^2 + 4^2 = r^2

9 + 16 = r^2

25 = r^2

r = 5

Note: we remark that 3,4 are the pythagorean numbers, so the radius could only be 5.

**The equation of the circle is:**

**(x - 2)^2 + (y)^2 = 25**