Question

The equation of the circle which has both the axes as its tangents and which passes through the point $(1,2)$

• A. $x^2+y^2-2x+2y-1=0$
• B. $x^2+y^2-2x+2y+1=0$
• C. $x^2+y^2-2x-2y+1=0$
• D. $x^2+y^2-2x-2y-1=0$

Answer 1

The correct option is C $x^2+y^2-2x-2y+1=0$

$R=a$

$(x-a{)}^2+(y-a{)}^2=a^2$

$(1,2)$

$(1-a{)}^2+(2-a{)}^2=a^2$

$\Rightarrow 1+a^2-2a+4+a^2-4a=a^2$

$=a^2-6a+5=0$

$\Rightarrow a=5,1$

$\therefore (x-1{)}^2+(y-1{)}^2=1^2$

$\Rightarrow x^2-2x+y^2-2y+1=0$.