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Решите задачу:

$a)\quad x^2+y^2-6x+8y=0\\\\(x-3)^2-9+(y+4)^2-16=0\\\\(x-3)^2+(y+4)^2=25\\\\okryznost\; ,\; \; centr\; v\; \; C(3;-4)\; ,\; R=5\\\\b)\quad 9x^2+16y^2=144\\\\ \frac{9x^2}{144}+\frac{16y^2}{144}=1\\\\\frac{x^2}{16}+\frac{y^2}{9}=1\\\\ellips\; ,\; \; centr\; v\; \; O(0;0)\; ,\; a=\sqrt{16}=4\; ,\; \; b=\sqrt9=3\\\\c^2=a^2-b^2=16-9=7\; ,\; \; c=\sqrt7\\\\F_1(c;0)\; ,\; \; F_2(-c;0)\\\\F_1(\sqrt7;0)\quad -\; \; pravuj\; fokys\\\\c)\quad pryamaya\; l:\; \; \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$

$C(3;-4)\; ,\; \; F_1(\sqrt7;0)\\\\ \frac{x-3}{\sqrt7-3}=\frac{y+4}{0+4}\\\\4(x-3)=(\sqrt7-3)(y+4)\\\\4x-12=(\sqrt7-3)y+4(\sqrt7-3)\\\\y=\frac{1}{\sqrt7-3} \cdot (4x-12-4(\sqrt7-3))\\\\y=\frac{4}{\sqrt7-3}\cdot (x-3-\sqrt7+3)\\\\y=\frac{4}{\sqrt7-3}\cdot x-\frac{4\sqrt7}{\sqrt7-3}\\\\y=\frac{4(\sqrt7+3)}{7-9}\cdot x- \frac{4\sqrt7(\sqrt7+3)}{7-9} \\\\y=-2(\sqrt7+3)\cdot x+2\sqrt7(\sqrt7+3)\\\\y=-2(\sqrt7+3)\cdot x+14+6\sqrt7$