A circle S passes through the point (0,1) and | vedclass

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(B) Let the circle be $x^2+y^2+2gx+2fy+c=0$ $\quad\quad(1)$The given circles are:$C_1: x^2+y^2-2x-15=0$ $\quad\quad(2)$$C_2: x^2+y^2-1=0$ $\quad\quad(

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$(B,C)$

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(B) Let the circle be $x^2+y^2+2gx+2fy+c=0$ $\quad\quad(1)$The given circles are:$C_1: x^2+y^2-2x-15=0$ $\quad\quad(2)$$C_2: x^2+y^2-1=0$ $\quad\quad(3)$Since $(1)$ is orthogonal to $(2)$,we have $2g_1g_2 + 2f_1f_2 = c_1+c_2$. Here $g_1=g, f_1=f, c_1=c$ and for $(2)$,$g_2=-1, f_2=0, c_2=-15$:$2(g)(-1) + 2(f)(0) = c-15 \Rightarrow -2g = c-15$ $\quad\quad(4)$Since $(1)$ is orthogonal to $(3)$,where $g_3=0, f_3=0, c_3=-1$:$2(g)(0) + 2(f)(0) = c-1 \Rightarrow c=1$Substituting $c=1$ into $(4)$:$-2g = 1-15 = -14 \Rightarrow g=7$Since the circle passes through $(0,1)$:$0^2+1^2+2g(0)+2f(1)+c=0$ $\Rightarrow 1+2f+1=0$ $\Rightarrow 2f=-2$ $\Rightarrow f=-1$The circle equation is $x^2+y^2+14x-2y+1=0$.Centre is $(-g, -f) = (-7, 1)$.Radius is $\sqrt{g^2+f^2-c} = \sqrt{7^2+(-1)^2-1} = \sqrt{49+1-1} = \sqrt{49} = 7$.Thus,$(B)$ and $(C)$ are correct.

$A$ 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 $7$
$(C)$ centre of $S$ is $(-7,1)$
$(D)$ centre of $S$ is $(-8,1)$

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$A$ 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)$ centre of $S$ is $(-7,1)$$(D)$ centre of $S$ is $(-8,1)$