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 [2014]

Question

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 [2014]

Smart Achievers · Accepted Answer

(2, 3)

Let the equation of circles be

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0 ...(i)$

It passes through (0, 1)

$∴$ $1 + 2 f + c = 0 ...(ii)$

Since equation (i) is orthogonal to circle ${(x - 1)}^{2} + y^{2} = 16$

$i.e. x^{2} + y^{2} - 2 x - 15 = 0$

and $x^{2} + y^{2} - 1 = 0$

$∴$ $2 g (- 1) + 2 f (0) = c - 15$

$\Rightarrow 2 g + c - 15 = 0 ...(iii)$

$and 2 g \cdot 0 + 2 f \cdot 0 = c - 1 \Rightarrow c = 1 ...(iv)$

$Solving (ii), (iii) and (iv), we get$

$c = 1, g = 7, f = - 1$

$∴$ $Required circle is x^{2} + y^{2} + 14 x - 2 y + 1 = 0,$ $with centre (- 7, 1) and radius 7$

$∴$ $(2) and (3) are correct options.$

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