Find the equation of a circle passing through the origin with center at $(3,4)$

To find the equation of a circle, we use the formula:

$(x – h)^2 + (y – k)^2 = r^2$

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

Given the center \( (3,4) \) and that the circle passes through the origin \( (0,0) \), we can find the radius \( r \) by calculating the distance between the center and the origin:

$r = \sqrt{(3 – 0)^2 + (4 – 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Thus, the equation of the circle is:

$(x – 3)^2 + (y – 4)^2 = 5^2$

or

$(x – 3)^2 + (y – 4)^2 = 25$

To determine the equation of the circle, we use the standard form:

$(x – h)^2 + (y – k)^2 = r^2$

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

The center is given as ( (3,4) ). By knowing the circle passes through the origin, the radius is the distance from ( (3,4) ) to ( (0,0) ):

$r = sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5$

Therefore, the equation of the circle is:

$(x – 3)^2 + (y – 4)^2 = 25$

The equation of a circle is:

$(x – h)^2 + (y – k)^2 = r^2$

Given center ( (3,4) ) and passing through the origin:

$r = sqrt{3^2 + 4^2} = 5$

So, the equation is:

$(x – 3)^2 + (y – 4)^2 = 25$

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