Question

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

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

Answer 1

Answer: (D)

$x ^ { 2 } + y ^ { 2 } - 6 x - 2 y + 1 = 0$

Answer 2

Joint equations of normals are $x ^ { 2 } - 3 x y - 3 x + 9 y = 0$

$x ( x - 3 y ) - 3 ( x - 3 y ) = 0$ $\Rightarrow$ $( x - 3 ) ( x - 3 y ) = 0$

$\therefore$ Given normals are $x - 3 = 0$ and $x - 3 y = 0$, which intersect at centre of circle whose coordinates are (3, 1).

The given circle is $C _ { 1 } = ( 3 , - 3 ) , \quad r _ { 1 } = 1$;

If the two circles touch externally, then $C _ { 1 } C _ { 2 } = r _ { 1 } + r _ { 2 }$

$\Rightarrow$ $4 = 1 + r _ { 2 } \Rightarrow r _ { 2 } = 3$

$\therefore$Equation of required circle is $( x - 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = ( 3 ) ^ { 2 }$

$x ^ { 2 } + y ^ { 2 } - 6 x - 2 y + 1 = 0$