What is the equation of a circle with center (5, 0) that passes through the point (1, 1)?

1 Answer
Shwetank Mauria
Jul 16, 2016

Equation of a circle is
#x^2+y^2-10x+8=0#

Explanation:

Let #(x,y)# be a point on circle. As it's distance from center #(5,0)# is equal to the distance between the point #(1,1)# and the center #(5,0)#. Hence

#(x-5)^2+(y-0)^2=(1-5)^2+(1-0)^2# or

#x^2-10x+25+y^2=(-4)^2+1^2# or

#x^2+y^2-10x+25=16+1# or

#x^2+y^2-10x+8=0#

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