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2 years ago

The circle below is centered at (4, -1) and has a radius of 3. What is its

Mathematics

1 answer:

denis23 [38]2 years ago

6 0

Answer:

Answers C and D are proably missing the y term. I'll assume they are:

C. (x-4)² + (y-1)² = 9

D. (x-4)² + (y-1)² = 3

A, is the equation that produces the drawn circle.

Step-by-step explanation:

See the attached image for explanation.

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A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago

DedPeter [7]

Answer:

The solution is (3/8, -7/8).

Step-by-step explanation:

y = −5x + 1

y = 3x − 2

Since the 2 expressions in x are both equal to y :

−5x + 1 = 3x - 2

-5x - 3x = -2 - 1

-8x = -3

x = 3/8.

So y = 3x - 2

= 3(3/8) - 2

= 9/8 - 2 = -7/8.

6 0

3 years ago

Translate triangle HIK 4 units to the right and 1 unit up.

Fiesta28 [93]

C because K is 1,-8 so if you go up 1, -7, and right 4,5. So you get k(5,-7)

4 0

2 years ago

3(q−7)=27 what is q mean in this equation

ArbitrLikvidat [17]

Probably Just The Variable

3 0

3 years ago

Read 2 more answers

Need help with this one please! Worth 30 points!!! :)

Kitty [74]

Answer:

y^2 +8y + 16

Step-by-step explanation:

6y^2 +2y +5 - (5y^2 -6y -11)

I distribute the minus sign

6y^2 +2y +5 - 5y^2 +6y +11

Then I put them vertical.

6y^2 +2y +5

-5y^2 +6y +11

------------------------

y^2 +8y + 16

This is in standard from since the exponential decreases.

5 0

3 years ago