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  • A circle S passes through the point (0,1) and is orthogonal to the circles <img alt="\mathrm{(x-1)^2+y^2=16 \ and \ x^2+y^2=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28x

A circle S passes through the point (0,1) and is orthogonal to the circles \mathrm{(x-1)^2+y^2=16 \ and \ x^2+y^2=1}, then radius of S is

Option: 1

7


Option: 2

6


Option: 3

8


Option: 4

4


Answers (1)

best_answer

Let equation of S is

\mathrm{x^2+y^2+2 g x+2 f y+c=0}       ...[i]

\mathrm{\text { Circle (i) cuts the circle }(x-1)^2+y^2=16}

 –2g = –15 + c

\mathrm{\text { Circle (i) also cuts } x^2+y^2=1 \text { orthogonally }}

0 = –1 + c ⇒ c = 1 and g = 7
Circle (i) passes through the point (0, 1)
2f + 1 + c = 0 ⇒ f = –1
Equation of the required circle is

\mathrm{\begin{aligned} & x^2+y^2+14 x-2 y+1=0 \\ & \therefore \quad \text { Radius }=\sqrt{49+1-1}=7 \end{aligned}}

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SANGALDEEP SINGH

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