The power of the point B(-1,1) with respect to the circle S x^2+y^2-2 x-4 y+3=0 is p . If the length of the tangent drawn from B to the circles S=0 is t , then the point (2,3) with respect to the circle S^ =0 having centre at (p, t^2 ) and passing through the origin.

Step-by-step solution

Given equation of circle array rlrl & & S & x^2+y^2-2 x-4 y+3=0 & p & =(-1)^2+(1)^2-2(-1)-4(1)+3 & & =1+1+2-4+3=3 & & t & = p t= 3 array Now, circle whose centre is (p, t^2 ) , i.e. (3,3) (x-3)^2+(y-3)^2=r^2 Since, this circle passes through (0,0) array ll & (0-3)^2+(0-3)^2=r^2 & r^2=9+9=18 array So, circle S^ will be (x-3)^2+(y-3)^2=18 Now, point (2,3) w.r.t. to circle aligned & (x-3)^2+(y-3)^2=18 is & =(2-3)^2+(3-3)^2-18 & =1-18=-17 < 0 aligned So, point (2,3) lies inside the circle S^ =0