Paragraph: Consider the family of circles x^2+y^2-2 x-2 y-8=0 passing through two fixed points A and B . Also S=0 is a circle of this family, the tangent to which at A and B intersect on the line x+2 y+5=0 . Question: If the circle x^2+y^2-10 x+2 y+c=0 is orthogonal to S=0 , then the value of c equals

Step-by-step solution

(i) We have (x^2+y^2-2 x-8 )-2 y=0 , which is the form of S+ L=0 On solving S=0 and L=0 where S x^2+y^2-2 x-8=0 and L y=0 Put y=0 x^2-2 x-8=0(x-4)(x+2)=0 x=-2 or 4 A (4,0) ; B (-2,0) Distance between A and B=6 (ii) Equation of A B which is chord of contact with respect to P (x₁, y₁ ) is xx ₁+ yy ₁-1 ( x + x ₁ )- ( y + y ₁ )-8=0 x (x₁-1 )+y (y₁- )- (x₁+ y₁+8 )=0 (1) Also equation of A B is x -axis i.e 0 x+1 y+0 c=0 (2) On comparing (1) & (2), we get x ₁-1 0 = y ₁- 1 = x ₁+ y ₁+8 0 =k say aligned & x ₁-1=0 & x ₁=1 ;