1.

Consider the family of circles x^(2)+y^(2)-2x-2ay-8=0 passingthrough two fixed points A and B . Also, S=0 is a cricle of this family, the tangent to which at A and B intersect on the line x+2y+5=0. If the circle x^(2)+y^(2)-10x+2y=c=0 is orthogonal to S=0, then the value of c is

Answer»

8
9
10
12

Solution :CIRCLE `x^(2)+y^(2)-10x+2y+C=0` is orthogonal to `x^(2)+y^(2)-2x-6y-8=0`.
So, by applying condition of orthogonal intersection, we get
`2(-5)(-1)+2(1) (-3)= c-8`
`:. C= 12`


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