Question

The equation of the circle which touches the circle $x^{2}+y^{2}-6 x+6 y+17=0$ externally and having the lines $x^{2}-3 x y-3 x+9 y=0$ as two normals, is

• A. $x^{2}+y^{2}-2 x+5 y-1=0$
• B. $x^{2}+y^{2}+2 x+3 y+1=0$
• C. $x^{2}+y^{2}-6 x-2 y+1=0$
• D. $x^{2}+y^{2}+4 x-3 y+3=0$

Answer 1

Answer: (C)

Equation of normals
$x^{2}-3 x y-3 x+9 y=0$
$\Rightarrow \, x(x-3 y)-3(x-3 y)=0$
$\Rightarrow \,(x-3 y)(x-3)=0$
So, equation of normals is $x-3 y=0$ and $x-3=0$ intersection of two normal is centre. So, $x-3 y=0$ and $x=3$.
$\Rightarrow \,x=3 y$
$ \Rightarrow \,3=3 y $
$ \Rightarrow \, y=1$
So, coordinate of centre $=(3,1)$
Now, equation of given circle
$x^{2}+y^{2}-6 x+6 y+17=0$
Centre $(3,-3)$ and radius $=\sqrt{3^{2}+(-3)^{2}-17}$
$r_{1}=\sqrt{1}=1$
Now, circles touches externally, so sum of radius = distance between centres $(3,1)$ and $(3,-3)$
$r_{1}+r_{2}=\sqrt{(3-3)^{2}+(1-(-3))^{2}}$
$1+r_{2}=4 \, r_{2}=3$
Equation of circle with centre $(3,1)$ and radius 3 units
$(x-3)^{2}+(y-1)^{2}=3^{2}$
$\Rightarrow \, x^{2}-6 x+9+y^{2}-2 y+1=9$
$\Rightarrow \,x^{2}+y^{2}-6 x-2 y+1=0$