Question

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)?

Answer 1

`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]}