Write an equation for a circle with a diameter that has endpointsat (-4, -7) and (-2 ,-5)

Mathematics

1 answer:

sergiy2304 [10]3 years ago

7 0

Answer:

(x + 3)^2 + (y + 6)^2 = (√2)^2 = 2

Step-by-step explanation:

The standard equation for a circle with center at (h, k) and radius r is

(x - h)^2 + (y - k)^2 = r^2. We need to find h, k and r for this particular circle.

The length of the radius is half the length of the diameter. Find the length of this diameter first, and than halve the result.

diameter^2 = 2^2 + 2^2, or diameter^2 = 8, or diameter = √8 = 2√2.

This is because x changes from -4 to -2 (+2) and y changes from -7 to -5 (+2).

Then the radius is r = 2√2.

The coordinates of the center are found using the midpoint formula:

-4-2

xcenter = -------- = -3

-7-5

ycenter = -------- = - 6

Inserting -3 for h, -6 for k and √2 for r, we get:

(x + 3)^2 + (y + 6)^2 = (√2)^2 = 2

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Sin^2(x/2)=sin^2x <br> Find the value of x

Veronika [31]

$\sin^2\dfrac x2=\sin^2x$
$\dfrac{1-\cos x}2=1-\cos^2x$
$1-\cos x=2-2\cos^2x$
$2\cos^2x-\cos x-1=0$
$(2\cos x+1)(\cos x-1)=0$
$\implies\begin{cases}2\cos x+1=0\\\cos x-1=0\end{cases}$ $\implies\begin{cases}\cos x=-\frac12\\\cos x=1\end{cases}$

The first case occurs in $0\le x$ for $x=\dfrac{2\pi}3$ and $x=\dfrac{4\pi}3$ . Extending the domain to account for all real $x$ , we have this happening for $x=\dfrac{2\pi}3+2n\pi$ and $\dfrac{4\pi}3+2n\pi$ , where $n\in\mathbb Z$ .

The second case occurs in $0\le x$ when $x=0$ , and extending to all reals we have $x=2n\pi$ for $n\in\mathbb Z$ , i.e. any even multiple of $\pi$ .