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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Correct answer: A. (a) Equation of circle passes through the point of intersections of given circles S_1 ≡ (x+3 )^2+ (y+2 )^2=25 and S_2 ≡ (x-2 )^2+ (y-3 )^2=25 is S_1+ S_2=0 ⇒ (1+ )x^2+ (1+ )y^2+ (6-4 )x+ (4-6 )y-12 (1+ )=0 or x^2+y^2+ ( 3-2 1+ )x+2 ( 2-3 1+ )y-12=0 … (i ) Cuts the another given circle (x+1 )^2+ (y-2 )^2=16 Orthogonally, We know condition of orthogonality 2g_1g_2+2f_1f_2=C_1+C_2 Hence, 2 ( 3-2 1+ )-4 ( 2-3 1+ )=-12-11 ⇒6-4 -8+12 =-23-23 ⇒31 =-21 ⇒ =- 2131 ∴ Centre of the circle (i) is (- 3+ 42311- 2131 ,- 2+ 63311- 2131)=(- 13510 ,- 12510)=(- 272 ,- 252)

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

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