The equation of the circle which intersects c | vedclass

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(D) Let the required circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$. Two circles ${x^2} + {y^2} + 2g_1x + 2f_1y + c_1 = 0$ and ${x^2} + {y^2} + 2g_2x +

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$3({x^2} + {y^2}) + 4(x + y) - 3 = 0$

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(D) Let the required circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$. Two circles ${x^2} + {y^2} + 2g_1x + 2f_1y + c_1 = 0$ and ${x^2} + {y^2} + 2g_2x + 2f_2y + c_2 = 0$ intersect at right angles if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. Applying this condition to the given circles: $(i) \; 2g(1/2) + 2f(1) = c + 3 \Rightarrow g + 2f = c + 3$ $(ii) \; 2g(1) + 2f(2) = c + 5 \Rightarrow 2g + 4f = c + 5$ $(iii) \; 2g(-7/2) + 2f(-4) = c - 9 \Rightarrow -7g - 8f = c - 9$ Subtracting $(i)$ from $(ii)$,we get $g + 2f = 2$. Substituting $g + 2f = 2$ into $(i)$,we get $2 = c + 3$,so $c = -1$. Now,$g + 2f = 2$ and $-7g - 8f = -1 - 9 = -10$,which simplifies to $7g + 8f = 10$. Solving $g + 2f = 2$ and $7g + 8f = 10$: Multiply the first by $4$: $4g + 8f = 8$. Subtracting this from $7g + 8f = 10$ gives $3g = 2$,so $g = 2/3$. Then $2/3 + 2f = 2$ $\Rightarrow 2f = 4/3$ $\Rightarrow f = 2/3$. Substituting $g=2/3, f=2/3, c=-1$ into the general equation: ${x^2} + {y^2} + 2(2/3)x + 2(2/3)y - 1 = 0$ ${x^2} + {y^2} + 4/3x + 4/3y - 1 = 0$ Multiplying by $3$: $3({x^2} + {y^2}) + 4x + 4y - 3 = 0$,which is $3({x^2} + {y^2}) + 4(x + y) - 3 = 0$.

The equation of the circle which intersects circles ${x^2} + {y^2} + x + 2y + 3 = 0$,${x^2} + {y^2} + 2x + 4y + 5 = 0$,and ${x^2} + {y^2} - 7x - 8y - 9 = 0$ at right angles is:

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