Determine the equation of a circle centered at (0, 0) and through the point of cutting the line 3x + 4y = 24 with the x axis?

1 Answer
Mar 11, 2018

Answer:

#x^2+y^2=64#

Explanation:

As the equation of #x#-axis i#y=0#, we can find the point of its intersection with the line #3x+4y=24# with the #x#-axis is obtained by putting #y-0#.

Then we have #3x+4xx0=24# or #3x=24# i.e. #x=8# and point of intersection is #(8,0)#.

As center is #(0,0)# and circle passes through #(8,0)#, its radius is #sqrt((8-0)^2+(0-0)^2)=sqrt64=8#

and hence equation of circle is

#(x-0)^2+(y-0)^2=8^2#

or #x^2+y^2=64#

graph{(x^2+y^2-64)(3x+4y-24)=0 [-16.83, 23.17, -9.36, 10.64]}