Find the equation of the circle passing through (2,5) and center at (4,2)?

To find the equation of the circle passing through (2,5) and center at (4,2), we first need to find the radius of the circle.

The distance between the center of the circle (4,2) and the point (2,5) is the radius of the circle. We can use the distance formula to find the radius:

r = √((x2 - x1)^2 + (y2 - y1)^2)
r = √((2 - 4)^2 + (5 - 2)^2)
r = √((-2)^2 + (3)^2)
r = √(4 + 9)
r = √13

Now that we have the radius, we can write the equation of the circle in the standard form:

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

where (h,k) is the center of the circle and r is the radius.

Plugging in the values we have:

(x - 4)^2 + (y - 2)^2 = 13

Therefore, the equation of the circle passing through (2,5) and center at (4,2) is:

(x - 4)^2 + (y - 2)^2 = 13

The equation of a circle with center (h, k) and radius r is given by
$$(x - h)^2 + (y - k)^2 = r^2$$
Given that the circle passes through (2,5) and it's center is (4,2)
Therefore, h=4 and k=2
So, the equation of the circle becomes
$$(x-4)^2+(y-2)^2=r^2$$
To find the value of r, we can substitute the given point (2,5)
$$(2-4)^2+(5-2)^2=r^2$$
$$(-2)^2+(3)^2=r^2$$
$$4+9=r^2$$
$$r^2=13$$
$$r=\sqrt{13}$$
Therefore the equation of the circle is
$$(x-4)^2+(y-2)^2=13$$