How to find the standard form of the equation of the specified circle given that it Is tangent to both axis and passes through ( 2 ,-1)?

1 Answer
May 25, 2017

Answer:

#x^2+y^2-10x+10y+25=0#

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

Explanation:

As the circle is tangent to both the axis, its centre is equidistant from both the axis

and as it passes through #(2,-1)#, it lies in fourth quadrant and coordinates of centre are of type #(a,-a)# and its equation is

#(x-a)^2+(y+a)^2=a^2#

or #x^2+y^2-2ax+2ay+a^2=0#

and as it passes through #(2,-1)#, we have

#2^2+(-1)^2-2axx2+2axx(-1)+a^2=0#

or #a^2-6a+5=0#

or #(a-5)(a-1)=0#

Hence #a=5# or #a=1#

and equation of circle could be

#x^2+y^2-10x+10y+25=0# or #x^2+y^2-2x+2y+1=0#

graph{(x^2+y^2-10x+10y+25)(x^2+y^2-2x+2y+1)=0 [-1.793, 8.207, -3.94, 1.06]}