Question

The equation of the circle touching both the lines $x+y=3,x-y=3$ and passing through $(0,0)$ is

• A. $x^2+6(1\pm \sqrt{2})y+y^2=0$
• B. $x^2-6(\sqrt{2}+1)y+y^2=0$
• C. $x^2+6(1\pm \sqrt{2})x+y^2=0$
• D. $x^2\pm 6(\sqrt{2}-1)x+y^2=0$

Answer 1

The correct option is C $x^2+6(1\pm \sqrt{2})x+y^2=0$

Let the equation of circle be $S=(x-a{)}^2+(y-b{)}^2=c^2$

$S$ passes through $(0,0)\Rightarrow a^2+b^2=c^2----\left(1\right)$

Given tangents : $x+y=3,x-y=3$

Distance of tangent from centre = radius

$\Rightarrow \left|\frac{a+b-3}{\sqrt{2}}\right|=\left|\frac{a-b-3}{\sqrt{2}}\right|=c$

$\Rightarrow |a+b-3{|}^2=|a-b-3{|}^2=(c\sqrt{2}{)}^2$

$\Rightarrow |a+b-3{|}^2=|a-b-3{|}^2=2c^2=2(a^2+b^2)$ from $\left(1\right)$

$\left(i\right)|a+b-3{|}^2=2(a^2+b^2)$

$\Rightarrow a^2+b^2+9+2(ab-3a-3b)=2a^2+2b^2$

$\Rightarrow a^2+b^2-2ab+6a+66-9=0---\left(2\right)$

$\left(ii\right)|a+b-3{|}^2=(a-b-3{)}^2$

$a^2+b^2+9+2ab-6b-6a=a^2+b^2+9-2ab+6b-6a$

$\Rightarrow 4ab-12b=0$

$\Rightarrow 4b(a-3)=0\Rightarrow a=3,b=0$

Put $a=3$ in $\left(2\right)\Rightarrow 9+b^2-6b+18+6b-9=0$

$\Rightarrow b^2+18=0$

Put $a=0$ in $\left(2\right)\Rightarrow b^2-9+6a=0$

$\Rightarrow b^2+6a-9=0\Rightarrow b=-3\pm 3\sqrt{2}=b=-3(1+\sqrt{2})$

$\Rightarrow (a,b)=(-3(1\pm \sqrt{2}),0)$

Equation of the circle $S=(x-a{)}^2+(y-b{)}^2=a^2$

$\Rightarrow x^2+a^2-2ax+y^2=a^2$ for $b=0$

$\Rightarrow x^2-2ax+y^2=0$

$\Rightarrow x^2+y^2\pm 6(1\pm \sqrt{2})x=0$