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Question

(A) Let the equation of the family of circles passing through the intersection of $S_1 \equiv (x+3)^2+(y+2)^2-25=0$ and $S_2 \equiv (x-2)^2+(y-3)^2-25

Vedclass · Accepted Answer

$\left(\frac{-27}{2}, \frac{-25}{2}\right)$

Vedclass · Answer

(A) Let the equation of the family of circles passing through the intersection of $S_1 \equiv (x+3)^2+(y+2)^2-25=0$ and $S_2 \equiv (x-2)^2+(y-3)^2-25=0$ be $S_1 + \lambda S_2 = 0$. Expanding this,we get $(1+\lambda)x^2 + (1+\lambda)y^2 + (6-4\lambda)x + (4-6\lambda)y + (13+13-25-25\lambda) = 0$. Dividing by $(1+\lambda)$,the equation becomes $x^2 + y^2 + \frac{6-4\lambda}{1+\lambda}x + \frac{4-6\lambda}{1+\lambda}y + \frac{26-25-25\lambda}{1+\lambda} = 0$. For the circle $x^2+y^2+2gx+2fy+c=0$ to cut $x^2+y^2+2g'x+2f'y+c'=0$ orthogonally,the condition is $2gg' + 2ff' = c+c'$. Here $g' = 1, f' = -2, c' = 1-4-16 = -19$. Substituting the values,$2(\frac{3-2\lambda}{1+\lambda})(1) + 2(\frac{2-3\lambda}{1+\lambda})(-2) = \frac{1-25\lambda}{1+\lambda} - 19$. Solving for $\lambda$,we get $\lambda = -\frac{21}{31}$. Substituting $\lambda$ back into the center formula $(-g, -f) = (-\frac{3-2\lambda}{1+\lambda}, -\frac{2-3\lambda}{1+\lambda})$,we obtain the center as $\left(-\frac{27}{2}, -\frac{25}{2}\right)$.

The centre of the circle passing through the points of intersection of the circles $(x+3)^2+(y+2)^2=25$ and $(x-2)^2+(y-3)^2=25$ and cutting the circle $(x+1)^2+(y-2)^2=16$ orthogonally is

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