The equation of a circle is in the format:

\(\displaystyle (x - h)^2 + (y- k)^2 = r^2\)

where \(\displaystyle (h,k)\) is the center and \(\displaystyle r\) is the radius.

Multiply two on both sides of the equation.

\(\displaystyle 2[\frac{(x-3)^2}{2}+\frac{(y-6)^2}{2} = 4]\)

The equation becomes:

\(\displaystyle (x-3)^2+(y-6)^2=8\)

The center is \(\displaystyle (3,6)\).

The radius is \(\displaystyle \sqrt8 = 2\sqrt2\).

The answer is:  \(\displaystyle c= (3,6), r= 2\sqrt2\)

To rewrite the given function as the equation of a circle in standard form, we must complete the square for x and y. This method requires us to use the following general form:

\(\displaystyle (a+b)^2=a^2+2ab+b^2\)

To start, we can complete the square for the x terms. We must halve the coefficient of x, square it, and add it to the first two terms:

\(\displaystyle (-2)^2=4\)

\(\displaystyle x^2-4x+4\)

Now, we can rewrite this as a perfect square, but because we added 4, we must subtract 4 as to not change the original function:

\(\displaystyle (x-2)^2-4\)

We do the same procedure for the y terms:

\(\displaystyle (-1)^2=1\)

\(\displaystyle y^2-2x+1\)

\(\displaystyle (y-1)^2-1\)

Rewriting our function, we get

\(\displaystyle (x-2)^2-4+(y-1)^2-1=11\)

Moving the constants to the right side, we get the function of a circle in standard form:

\(\displaystyle (x-2)^2+(y-1)^2=16\)

Comparing to

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

we see that the radius of the circle is

\(\displaystyle \sqrt{16}=4\)

Notice that the radius is a distance and can therefore never be negative.

When identifying the center of a circle, take the opposite sign of each value connected to x and y. 

\(\displaystyle (x+4)^2+(y-2)^2=49\ compare\ with\ the\ general\ equation\ of\ a\ circle: (x-h)^2+(y-k)^2=r^2\)

\(\displaystyle Changing\ signs\ for\ each\ gives\ us\ a\ center\ of: (-4,2)\)

\(\displaystyle To\ find\ the\ radius\ we\ set\ r^2=49,\)

\(\displaystyle take\ square\ root\ of\ both\ sides\ to\ get\ r=7.\)

The equation given represents a circle.

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

\(\displaystyle (h,k)\) represents the center, and \(\displaystyle r\) is the radius.

\(\displaystyle h=6 , k = -7\)

The center is at:  \(\displaystyle (6,-7)\)

Set up an equation to solve the radius.

\(\displaystyle r^2=2\)

The radius is:  \(\displaystyle r=\sqrt{2}\)

The answer is:  \(\displaystyle (6,-7); \sqrt2\)

Write the standard form for a circle.

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

The circle is centered at:  \(\displaystyle (h,k)\)

The radius is:  \(\displaystyle r=2\)

Substitute all the known values into the formula.

\(\displaystyle (x-(-1))^2+(y-2)^2=2^2\)

Simplify this equation.

The answer is:  \(\displaystyle (x+1)^2+(y-2)^2=4\)

First, we need to become familiar with the standard form of a hyperbolic equation:

\(\displaystyle \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\)

The center is always at \(\displaystyle (h,k)\). This means that for this problem, the numerators of each term will have to contain \(\displaystyle (x+4)^{2}\) and \(\displaystyle (y-3)^{2}\).

To determine if a hyperbola opens vertically or horizontally, look at the sign of each variable. A vertical parabola has a positive \(\displaystyle y^{2}\) term; a horizontal parabola has a positive \(\displaystyle x^{2}\) term. In this case, we need a vertical parabola, so the \(\displaystyle y^{2}\) term will have to be positive.

(NOTE: If both terms are the same sign, you have an ellipse, not a parabola.)

The asymptotes of a parabola are always found by the equation \(\displaystyle y=\pm\, \frac{b}{a}\, x\), where \(\displaystyle b\) is found in the denominator of the \(\displaystyle y^{2}\) term and \(\displaystyle a\) is found in the denominator of the \(\displaystyle x^{2}\) term. Since our asymptotes are \(\displaystyle y=\pm\, \frac{4}{3}\, x\), we know that \(\displaystyle b\) must be 4 and \(\displaystyle a\) must be 3. That means that the number underneath the \(\displaystyle y^{2}\) term has to be 16, and the number underneath the \(\displaystyle x^{2}\) term has to be 9.

The standard equation of a hyperbola is:

\(\displaystyle \small \frac{y^2}{a^2}-\frac{x^2}{b^2}=1\)

In order to express the given hyperbolic function in standard form, we must write it in one of the following two ways:

\(\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)

\(\displaystyle \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\)

From our formulas for the standard form of a hyperbolic equation above, we can see that the term on the right side of the equation is always 1, so we must divide both sides of the given equation by 52, which gives us:

\(\displaystyle \frac{13(y-7)^2}{52}-\frac{13(x+6)^2}{52}=1\)

Simplifying, we obtain our final answer in standard form:

\(\displaystyle \frac{(y-7)^2}{4}-\frac{(x+6)^2}{4}=1\)

The correct definition of hyperbolic sine is:

\(\displaystyle \sinh(x)= \frac{e^x-e^-^x}{2}\)

Therefore, by multiplying 2 by both sides we get the following answer,

\(\displaystyle 2\sinh(x)=e^x-e^-^x\)

Find the values of hyperbolic sine and cosine when x is zero.  According to the properties:

\(\displaystyle \sinh(0)=\frac{e^x-e^{-x}}{2}=\frac{e^0-e^{-0}}{2}=\frac{1-1}{2}=0\)

\(\displaystyle \cosh(0)= \frac{e^x+e^{-x}}{2}=\frac{e^0+e^{-0}}{2}=\frac{1+1}{2}=\frac{2}{2}=1\)

Therefore:

\(\displaystyle \cosh(0)-\sinh(0)=1-0=1\)