以标准形式重写圆的方程:x 2 + y 2 + 6x – 4y – 12 = 0
圆锥这个词的意思是与圆锥有关的东西。圆锥是一个 3D 图形,现在成像一个平面,任何通过圆锥的平面,两者之间的相交曲线会产生不同的圆锥截面。圆锥曲线一般有 3 种基本类型,椭圆、抛物线和双曲线。为什么没有提到圈子?这是因为圆是一种特殊类型的椭圆,具有相等的 x 和 y 系数。描述所有圆锥曲线的方程为,ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0,其中 a、b 不能同时为 0。
上述方程是圆锥曲线的一般方程。基于 a、b 和 h 的值,该等式说明了它所代表的圆锥截面的类型。
圆圈
圆被定义为与平面上的固定点等距的每个点的轨迹。在用于表示任何类型圆锥的一般方程中,也可以形成圆方程。如果 a = b 且 h = 0,
然后得到一个圆,方程将变为ax 2 + ay 2 + 2gx + 2fy +c = 0。这就是圆的一般方程。也可以改写如下,
x 2 + y 2 + 2gx/a + 2fy/a + c/a = 0
或者,(x + g/a) 2 + (y + f/a) 2 = (g/a) 2 + (f/a) 2 – c/a ⇢ (1)
这个圆的中心在 (-g/a, -f/a) 并且圆的半径是
如果圆的中心在 (0, 0) 原点。圆的方程将是 x 2 + y 2 = 1
以标准形式写出圆的方程 x 2 + y 2 + 6x – 4y – 12 = 0
解决方案:
In the question, It is asked to write the equation of the circle in standard form. The equation given is x2 + y2 + 6x – 4y – 12 = 0. From the equation itself, it can be said that this is an equation of a circle since the coefficients of x and y are equal. This equation can be rewritten as follows:
(x + 3)2 – 9 + (y – 2)2 – 4 – 12 = 0
Or, (x + 3)2 + (y – 2)2 = 25
Or, (x + 3)2 + (y – 2)2 = 52
示例问题
问题 1:写出标准形式的圆方程 x 2 + y 2 – 6x – 8y – 1 = 0
解决方案:
Compare this equation with general equation of circle which is ax2 + ay2 + 2gx + 2fy + c = 0
After comparing, a = 1, g = -3, f = – 4, and c = -1
Now that the values are obtained, one can easily form the standard equation of the circle by plugging the values into equation (1),
(x – 3/1)2 + (y – 4/1)2 = (-3/1)2 + (-4/1)2 + (-1/1)
Or, (x – 3)2 + (y – 4)2 = 9 + 16 -1
Or, (x – 3)2 + (y – 4)2 = 24
问题 2:写出以原点为原点,半径为 5 个单位的圆的方程。
解决方案:
Center at origin means the coordinate is (0, 0). The standard equation of a circle is (x – a)2 + (y – b)2 = r2.
Where (a, b) is the center of the circle and r being its radius.
Here, in the question center is (0, 0) and the radius is 5 units. therefore the equation of the circle will be,
(x – 0)2 +(y – 0)2 = 52
x2 + y2 = 25
问题 3:写出圆心在 (5,-3) 且半径为 4 个单位的圆的方程。
解决方案:
Center at (5, -3). The standard equation of a circle is (x – a)2 + (y – b)2 = r2.
Therefore the equation of the circle will be (x – 5)2 + (y – (-3))2 = 42
Or, (x – 5)2 + (y + 3)2 = 16
问题 4:找到由方程 x 2 + y 2 – 2x + 8y – 8 = 0 定义的圆的中心和半径。
解决方案:
The above equation can be rewritten as x2 – 2x + 1 + y2 + 8y + 16 = 8 + 1 +16
Or, (x – 2)2 + (y + 4)2 = 25
Or, (x – 2)2 + (y + 4)2 = 52