How do you write the standard form of the equation of the circle that is tangent to x=-2 and has its center at (2, -4)?

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Answer 1 · 2016-06-14T12:53:21

Equation of circle is #x^2+y^2-4x+8y+4=0#

Distance from a point #(m,n)# to a line #ax+by+c=0# is given by

#(am+bn+c)/sqrt(a^2+b^2)#

Hence, length of perpendicular from #(2,-4)# to #x=-2# or #x+2=0# is #(2+2)/1=4#

Hence #4# is radius and as center is #(2,-4)#, equation of circle will be

#(x-2)^2+(y+4)^2=16# or

#x^2-4x+4+y^2+8y+16=16#

or #x^2+y^2-4x+8y+4=0#