Discussion Forum

1)

The lines represented by $5x^{2}-xy-5x+y=0$ are normals to a circle S=0 .If this circle touches  the circle  

$S'=  x^{2}+y^{2}-2x+2y-7=0$ externally , then  the equation of the chord of contact  of centre of S'=0 with respect to S=0 is 

Answer:

Option A

Explanation:

 We have,

$5x^{2}-xy-5x+y=0$  are normal to circle S=0

$\therefore$ Centre of circle S=0  is point of intersection of line

              $5x^{2}-xy-5x+y=0$

           (x-1)(5x-y)=0

x=1, 5x-y=0

$\therefore$  Centre (1,5)

Centre of circle        S'= $x^{2}+y^{2}-2x+2y-7=0$  is (1,-1)

  and radius =3

 $SS'=\sqrt{(1-1)^{2}+(5+1)^{2}}=6$

 $\therefore$       r= SS'-3=6-3=3

Equation of circle 

 S= $(x-1)^{2}+(y-5)^{2}= (3)^{2}$

 = $x^{2}+y^{2}-2x-10y+17=0$

 Equation of chord of contact at (1,-1) to S=0 is

 $x-y-2\frac{(x-1)}{2}-10\frac{(y-1)}{2}+17=0$

 x-y-x-1-5y+5+17=0

 $\Rightarrow$     -6y+21=0

$\Rightarrow$       2y-7=0