Question

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

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

Answer 1

Answer: (C)

Equation of normals
$x^2-3xy-3x+9y=0$
$\Rightarrow x(x-3y)-3(x-3y)=0$
$\Rightarrow (x-3y)(x-3)=0$
So, equation of normals is $x-3y=0$ and $x-3=0$ intersection of two normal is centre. So, $x-3y=0$ and $x=3$.
$\Rightarrow x=3y$
$\Rightarrow 3=3y$
$\Rightarrow y=1$
So, coordinate of centre $=(3,1)$
Now, equation of given circle
$x^2+y^2-6x+6y+17=0$
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 $(3,-3)$
$r_1+r_2=\sqrt{(3-3{)}^2+(1-(-3){)}^2}$
$1+r_2=4r_2=3$
Equation of circle with centre $(3,1)$ and radius 3 units
$(x-3{)}^2+(y-1{)}^2=3^2$
$\Rightarrow x^2-6x+9+y^2-2y+1=9$
$\Rightarrow x^2+y^2-6x-2y+1=0$