A circle S passes through the point (0, 1) and is orthogonal to the circles `(x -1)^2 + y^2 = 16` and `x^2 + y^2 = 1`. Then(A) radius of S is 8(B) radius of S is 7(C) center of S is (-7,1)(D) center of S is (-8,1)A. radius of S is 8B. radius of S is 7C. centre of S is `( -7,1)`D. centre of S is `( -8,1)`

A circle S passes through the point (0, 1) and is orthogonal to the ci

1.	A circle S passes through the point (0, 1) and is orthogonal to the circles `(x -1)^2 + y^2 = 16` and `x^2 + y^2 = 1`. Then(A) radius of S is 8(B) radius of S is 7(C) center of S is (-7,1)(D) center of S is (-8,1)A. radius of S is 8B. radius of S is 7C. centre of S is `( -7,1)`D. centre of S is `( -8,1)`
Answer» Correct Answer - 2,3 Given circles `S_(1) : x^(2)+y^(2)-2x-15=0` and `S_(2) : x^(2)+y^(2)-1=0` Center of the circle which intersects above two circles orthogonally lies on the radical axis of the circles which is `S_(1)-S_(2) =0` or `x+7=0` Let the centre of the required circle be `C (-7,k)` Circle passes throught the point `A (0,1)`. `:. ` radius , `sqrt(7^(2)+(k-1)^(2))` Also, radius `=` length of the tangent from C to the any one of the given circles. `:. r=sqrt(7^(2)+k^(2)-1)` Comparing , we get `7^(2)+(k-1)^(2)=7^(2)+k^(2)-1` or `-2k+1= -1` or `k=1` `:. r=7`

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