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3 years ago

9

What is the radius of a circle with the equation x^2+y^2+6x-2y+3?

2 answers:

atroni [7]3 years ago

8 0

Answer:

r = √13

Step-by-step explanation:

Starting with x^2+y^2+6x-2y+3, group like terms, first x terms and then y terms: x^2 + 6x + y^2 -2y = 3. Please note that there has to be an " = " sign in this equation, and that I have taken the liberty of replacing " +3" with " = 3 ."

We need to "complete the square" of x^2 + 6x. I'll just jump in and do it: Take half of the coefficient of the x term and square it; add, and then subtract, this square from x^2 + 6x: x^2 + 6x + 3^2 - 3^2. Then do the same for y^2 - 2y: y^2 - 2y + 1^2 - 1^2.

Now re-write the perfect square x^2 + 6x + 9 by (x + 3)^2. Then we have x^2 + 6x + 9 - 9; also y^2 - 1y + 1 - 1. Making these replacements:

(x + 3)^2 - 9 + (y - 1)^2 -1 = 3. Move the constants -9 and -1 to the other side of the equation: (x + 3)^2 + (y - 1)^2 = 3 + 9 + 1 = 13

Then the original equation now looks like (x + 3)^2 + (y - 1)^2 = 13, and this 13 is the square of the radius, r: r^2 = 13, so that the radius is r = √13.

Komok [63]3 years ago

5 0

Answer:

2.646

Step-by-step explanation:

Complete the square, put the equation in standard form, then evaluate the radius.

... x² + y² + 6x - 2y +3 = 0 . . . . given equation in general form

... x² +6x + 9 +y² -2y +1 = -3 +9 +1 . . . . add the squares of half the coefficients of the x- and y-terms; subtract 3

... (x +3)² +(y -1)² = 7 . . . . standard form equation for a circle

The term on the right, 7, is the square of the radius. so the radius is ...

... r = √7 ≈ 2.646

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<u>Given</u>:

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<u>To Find</u>:

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2 years ago

URGENT! WILL GIVE BRAINLIEST!!

IgorC [24]

Answer:

Option C) $a_{n}=m-b+m(n-1)$ for $n={\{1,2,3,...}\}$ is not the correct way to define the given infinite sequence

${\{m+b,2m+b,3m+b,4m+b,...}\}$

Step-by-step explanation:

Given infinite sequence is ${\{m+b,2m+b,3m+b,4m+b,...}\}$

Option B) $a_{n}=m-b+m(n-1)$ for $n={\{1,2,3,...}\}$ is not the correct way to define the given infinite sequence ${\{m+b,2m+b,3m+b,4m+b,...}\}$

Now verify $a_{n}=m-b+m(n-1)$ for $n={\{1,2,3,...}\}$ is true for the given infinite sequence

That is put n=1,2,3,.. in the above function

$a_{n}=m-b+m(n-1)$

When n=1, $a_{1}=m-b+m(1-1)$

$=m-b+0$

$a_{1}=m-b\neq m+b$

When n=2, $a_{2}=m-b+m(2-1)$

$=m-b+m$

$a_{2}=2m-b\neq 2m+b$

When n=3, $a_{3}=m-b+m(3-1)$

$=m-b+2m$

$a_{3}=3m-b\neq 3m+b$

and so on.

Therfore $a_{n}=m-b+m(n-1)$ for $n={\{1,2,3,...}\}$ is not the correct way to define the given infinite sequence

${\{m+b,2m+b,3m+b,4m+b,...}\}$

Therefore option C) is correct

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3 years ago