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?

1 Answer
Shwetank Mauria
Apr 11, 2017

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]}

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