Find the equation of a circle, which passes through origin and has #x#-intercept as #3# and #y#-intercept as #4#? What would have been the equation, if intercepts are reversed?

graph{(x^2+y^2-3x-4y)(x^2+y^2-0.01)((x-3)^2+y^2-0.01)(x^2+(y-4)^2-0.01)=0 [-3.77, 6.23, -0.6, 4.4]}

graph{(x^2+y^2-4x-3y)(x^2+y^2-0.01)((x-4)^2+y^2-0.01)(x^2+(y-3)^2-0.01)=0 [-3.667, 6.33, -0.84, 4.16]}

To find the equation of a circle passing through the origin with x-intercept 3 and y-intercept 4, we can use the general equation of a circle:

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

Where (h, k) is the center of the circle and r is the radius. Since the circle passes through the origin, the center is (0, 0).

Now, let's find the radius using the intercepts:

The x-intercept is (3, 0) and the y-intercept is (0, 4). The distance between these two points (radius) can be found using the distance formula:

[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

[ r = \sqrt{(3 - 0)^2 + (0 - 4)^2} ] [ r = \sqrt{3^2 + (-4)^2} ] [ r = \sqrt{9 + 16} ] [ r = \sqrt{25} ] [ r = 5 ]

So, the radius is 5. Now, we can substitute the values into the equation:

[ (x - 0)^2 + (y - 0)^2 = 5^2 ] [ x^2 + y^2 = 25 ]

Therefore, the equation of the circle passing through the origin with x-intercept 3 and y-intercept 4 is ( x^2 + y^2 = 25 ).

If the intercepts are reversed, meaning x-intercept is 4 and y-intercept is 3, the equation would remain the same because the circle would still have the same radius and pass through the origin. So, the equation would still be ( x^2 + y^2 = 25 ).