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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