A circle has a center that falls on the line #y = 2x +7 # and passes through #(4 ,7 )# and #(1 ,2 )#. What is the equation of the circle?

1 Answer
May 17, 2016

Equation of the circle is #169x^2+169y^2+130x-2106y+3237=0#

Explanation:

The circle passes through line #y=2x+7# and it also passes through #(4,7)# and #(1,2)#. As such it should be equidistant from these two points. Hence,

#(x-4)^2+(y-7)^2=(x-1)^2+(y-2)^2#

or #x^2-8x+16+y^2-14y+49=x^2-2x+1+y^2-4y+4#

or #-8x+2x-14y+4y+16+49-1-4=0#

or #-6x-10y+60=0# but as #y=2x+7#

#-6x-10(2x+7)+60=0# or #-26x-10=0#

or #x=-10/26=-5/13# and #y=2xx-5/13+7=-10/13+7=81/13#

Hence center of circle is #(-5/13,81/13)# and radius of circle will its distance with #(1,2)# or

#r^2=(1+5/13)^2+(2-81/13)^2=(18/13)^2+(-55/13)^2=(324+3025)/169=3349/169#

And equation of circle is

#(x+5/13)^2+(y-81/13)^2=3349/169# multiplying by #169#

#(13x+5)^2+(13y-81)^2=3349# or

#169x^2+130x+25+169y^2-2106y+6561=3349#

or #169x^2+169y^2+130x-2106y+3237=0#