A circle having centre at C is made to pass through the point P(1,2) , touching the straight lines 7 x-y=5 and x+y+13=0 at A and B respectively, then

Step-by-step solution

Clearly, the point lies on 7 x-y=5 Also, centre of the circle must lie on the bisectors of the lines x+y+13=0 and 7 x-y-5=0 given by x+y+13 2 = 7 x-y-5 50 x-3 y=35 and 3 x+y+15=0 Let (h, k) be the centre of the circle, then h-3 k=35 (1) and 3 ~h + k =15 (2) Clearly C B is perpendicular B P k-2 h-1 7=-1 h+7 k-15=0 (3) On solving, we get centres as C₁ (29-2) and C₂ (-6,3) r₁^2=800 and r₁^2=50 smaller circle has radius = 50 Therefore area of quadrilateral ACBP =2 [ 1 2 50 200 ] sq. units