Find the equation of a circle whose center is #(5,-1)# and tangent is #y#-axis?

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Answer 1 · 2018-01-24T13:11:23

#(x-5)^2+(y+1)^2=25#

or #x^2+y^2-10x+2y+1=0#

The circle has its center at #(5,-1)# and #y#-axis i.e. #x=0# is the tangent.

Hence, the radius is the length of perpendicular from #(5,-1)# to #x=0# and this distance is #5#, the abscissa of the point.

Hence, equation of circle is

#(x-5)^2+(y+1)^2=5^2#

i.e. #x^2-10x+25+y^2+2y+1=25#

i.e. #x^2+y^2-10x+2y+1=0#

graph{((x-5)^2+(y+1)^2-25)((x-5)^2+(y+1)^2-0.03)=0 [-7.42, 12.58, -5.72, 4.28]}