Write the equation of the circle in general form. Show all of your work

Mathematics

2 answers:

finlep [7]3 years ago

8 0

<h2>1. What are the center and equation of the circle?</h2>

In this exercise we have the general form of the equation of a circle, which is given by:

$x^2+y^2+18x+4y+49=0$

The ordinary equation of a circle is:

$(x-h)^2+(y-k)^2=r^2 \\ \\ Developing \ this \ equation: \\ \\ x^2+y^2-2hk-2ky+h^2+k^2-r^2=0$

But this can be written as:

$x^2+y^2+Dx+Ey+F=0 \\ \\ D=-2h, \ E=-2k, \ and \ F=h^2+k^2-r^2$

From the original equation we know that:

$D=18, \ E=4, \ and \ F=49 \\ \\ So: \\ \\ h=-\frac{18}{2}=9, \ k=-\frac{4}{2}=2, \ and \ r=\sqrt{9^2+2^2-49}=6$

Finally:

$\boxed{CENTER: \ (h,k)=(9,2)} \\ \\ \boxed{RADIUS: \ 6}$

<h2>2. Write the equation of the circle in general form.</h2>

We need to write this equation in the form:

$x^2+y^2+Dx+Ey+F=0$

Since:

$D=-2h, \ E=-2k, \ and \ F=h^2+k^2-r^2$

Then we need to find the center and radius in order to get D, E and F then. From the graph, it is easy to know that the center is $(h,k)=(-3,4)$ and the radius is 2. Therefore:

$D=-2(-3)=6, \ E=-2(4)=-8, \ and \ F=(-3)^2+(4)^2-(2)^2=21$

Finally, the general form of the equation is:

$\boxed{x^2+y^2+6x-8y+21=0}$

<h2>3. Write the equation of the parabola</h2>

A parabola is the set of all points in a plane that are equidistant from a fixed line called the directrix and a fixed point called the focus that does not lie on the line. Since the directrix is $x=2$ then this is a horizontal axis. So the standard for of the equation of a parabola that matches this form is: