What is the equation of the circle with its center at (0,0) that passes through (3,4).

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

Now we have the center of the circle at (0,0). It passes through the point (3,4).

So the radius is sqrt...

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

Now we have the center of the circle at (0,0). It passes through the point (3,4).

So the radius is sqrt ( 3^2 + 4^2)

=> sqrt ( 16 + 9)

=> sqrt 25

=> 5

Substituting the coordinates of the center (0,0) and the radius 5, we get (x - 0)^2 + (y - 0)^2 = 5^2

=> x^2 + y^2 = 25

The equation of the circle is x^2 + y^2 = 25

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Given that the center of the circle is the point (0,0).

We know that the equation of the circle is given by :

(x-a)^2 + (y-b)^2 = r^2 where (a,b) is the center and r is the radius.

==> x^2 + y^2 = r^2

Now we will determine the radius.

Since the point (3,4) is on the circle, then the distance between the point and the center is the radius.

==> r = sqrt( 3^2 + 4^2) = sqrt25 = 5

Then, the radius of the circle is 5 units.

==> Them the equation of the circle is:

x^2 + y^2 = 25

Approved by eNotes Editorial Team