Consider the circle #x^2+y^2-2x-14y+25=0#. For what values of m is the line y=mx tangent to this circle?

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Answer 1 · 2018-05-20T12:39:16

#m=-4/3# or #m=3/4#

#x^2+y^2-2x-14y+25=0#

#x^2-2x+y^2-14y=-25#

#x^2-2x+1+y^2-14y+49=-25+1+49#

#(x-1)^2+(y-7)^2=25#

Therefore, the centre is #(1,7)# and the radius is 5

For a line to be tangent to the circle, then we need to find using the perpendicular distance formula

Your line: #y=mx# --> #y-mx=0#
Your (x,y) --> (1,7)

#(abs(y-mx))/sqrt(a^2+b^2)=5#

#(abs(7-m))/sqrt(1^2+m^2)=5#

#(7-m)/sqrt(1^2+m^2)=5# or #(-1)(7-m)/sqrt(1^2+m^2)=5#
(you'll get the same answer in both cases so I'm just going to do one of them)

#(7-m)/sqrt(1^2+m^2)=5#

#7-m=5sqrt(1+m^2)#

#(7-m)^2=25(1+m^2)#

#49-14m+m^2=25+25m^2#

#24m^2+14m-24=0#

#12m^2+7m-12=0#

#(3m+4)(4m-3)=0#

#m=-4/3# or #m=3/4#