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перехваты (x+2)/(x-1)

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центр (x−4)²+(y+2)²=25

Solution

(4, −2)

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(x−4)²+(y+2)²=25

Уравнение окружности

(x−a)²+(y−b)²=r² представляет собой уравнение окружности с радиусом r, расположеном в центре точки (a, b)

Перепишите (x−4)²+(y+2)²=25 в виде стандартного уравнения окружности

(x−4)²+(y−(−2))²=5²

Следовательно, свойства окружности:

(a, b)=(4, −2), r=5

А центом является:

(4, −2)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

перехваты (x+2)/(x-1)

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