Question

1. (HARDER] Consider the equation x2 + y2 + Ax + By + C = 0, where A, B, and care constants. Show that its graph is a circle if A2 + B2 > 4C. Find its centre and radius. What happens if AP + B2 < 4C? 2. Which of the following rules define functions? (a) The rule that assigns to each person in a classroom his or her height (b) The rule that assigas to each mother her youngest child alive today (C) The rule that assigns the perimeter of a rectangle to its area. (d) The rule that assigns the surface area of a spherical ball to its volume. (e) The rule that assigns the pair of numbers (x + 3, y) to the pair of numbers (x, y).​

Answer 1

Answer:

15+-282-

Step-by-step explanation:

1272+07-_3891

Answer 2

SOLUTION

GIVEN

Consider the equation x² + y² + Ax + By + C = 0, where A, B, and care constants.

TO DETERMINE

1. Show that its graph is a circle if A² + B² > 4C.

2. Find its centre and radius.

3. What happens if A² + B² < 4C

EVALUATION

Here the given equation is

x² + y² + Ax + By + C = 0

We simplify it as below

$\displaystyle \sf \: {x}^{2} + {y}^{2} + Ax + By + C = 0$

$\displaystyle \sf \implies {x}^{2} + {y}^{2} + 2.x. \frac{A}{2} + 2.y . \frac{B}{2} + C = 0$

$\displaystyle \sf \implies {\bigg(x + \frac{A}{2} \bigg)}^{2} + {\bigg(y + \frac{B}{2} \bigg)}^{2} - \frac{ {A}^{2} }{4} - \frac{ {B}^{2} }{4} + C = 0$

$\displaystyle \sf \implies {\bigg(x + \frac{A}{2} \bigg)}^{2} + {\bigg(y + \frac{B}{2} \bigg)}^{2} = \frac{ {A}^{2} }{4} + \frac{ {B}^{2} }{4} - C$

$\displaystyle \sf \implies {\bigg(x + \frac{A}{2} \bigg)}^{2} + {\bigg(y + \frac{B}{2} \bigg)}^{2} = \frac{ {A}^{2} + {B}^{2} -4 C}{4}$

$\displaystyle \sf \implies {\bigg(x + \frac{A}{2} \bigg)}^{2} + {\bigg(y + \frac{B}{2} \bigg)}^{2} = { \bigg( \frac{ \sqrt{ {A}^{2} + {B}^{2} -4 C}}{2} \bigg)}^{2}$

If the above equation represents a circle

Then radius > 0

$\displaystyle \sf \implies \frac{ \sqrt{ {A}^{2} + {B}^{2} -4 C}}{2} > 0$

$\displaystyle \sf \implies {A}^{2} + {B}^{2} -4 C > 0$

$\displaystyle \sf \implies {A}^{2} + {B}^{2} > 4 C$

2.

$\displaystyle \sf Centre \: of \: the \: circle = \bigg( - \frac{A}{2} , - \frac{B}{2} \bigg)$

$\displaystyle \sf Radius \: of \: the \: circle = \frac{ \sqrt{ {A}^{2} + {B}^{2} -4 C}}{2}$

3. If A² + B² < 4C

Then the equation becomes imaginary . This type of circle is called an imaginary circle.

More precisely the equation x² + y² + Ax + By + C = 0 does not represent any real circle. Hence it is not possible to draw the circle.

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