The centre of the circle,which cuts orthogona | vedclass

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(A) Let the required circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ $... (i)$The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 +

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$(3, 2)$

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(A) Let the required circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ $... (i)$The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Applying this to circle $(i)$ and the given circles:$1$) $2g(1) + 2f(8.5) = c + 4 \implies 2g + 17f = c + 4$ $... (ii)$$2$) $2g(3.5) + 2f(3) = c + 11 \implies 7g + 6f = c + 11$ $... (iii)$$3$) $2g(-0.5) + 2f(11) = c + 3 \implies -g + 22f = c + 3$ $... (iv)$Subtracting $(iii)$ from $(ii)$: $(2g - 7g) + (17f - 6f) = (c - c) + (4 - 11) \implies -5g + 11f = -7$ $... (v)$Subtracting $(iv)$ from $(iii)$: $(7g - (-g)) + (6f - 22f) = (c - c) + (11 - 3) \implies 8g - 16f = 8 \implies g - 2f = 1 \implies g = 2f + 1$ $... (vi)$Substituting $(vi)$ into $(v)$: $-5(2f + 1) + 11f = -7 \implies -10f - 5 + 11f = -7 \implies f = -2$.Substituting $f = -2$ into $(vi)$: $g = 2(-2) + 1 = -3$.The centre of the circle is $(-g, -f) = (-(-3), -(-2)) = (3, 2)$.

The centre of the circle,which cuts orthogonally each of the three circles $x^2 + y^2 + 2x + 17y + 4 = 0$,$x^2 + y^2 + 7x + 6y + 11 = 0$,and $x^2 + y^2 - x + 22y + 3 = 0$,is

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