Question

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?

Answer 1

Equation of circle is `x^2+y^2-3x-4y=0`. If intercepts are reversed equation would be `x^2+y^2-4x-3y=0`.

Explanation:

It is assumed that intercept on `x`-axis is `3` and intercept on `y`-axis is `4`. As such the circle passes through `(3,0)`, `(4,0)` and also `(0,0)`.

Let the equation of circle be `x^2+y^2+2gx+2fy+c=0`

As circle passes through `(0,0)`, we have

`0^2+0^2+2gxx0+2fxx0+c=0` or `c=0`

as it passes through `(3,0)`, we have

`3^2+0^2+2gxx3+2fxx0+0=0` or `6g=-9` or `g=-3/2`

as it also passes through `(0,4)`, we have

`0^2+4^2+2gxx0+2fxx4+0=0` or `8f=-16` or `f=-2` and hence

Equation of circle is `x^2+y^2-3x-4y=0`

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]}

Had the `x`-intercept been `4` and `y`-intercept been `3`, the equation would have been `x^2+y^2-4x-3y=0` and it appears as

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]}