Question

Given the general form of the equation of a circle
x2 + 10x + y2 – 4y + 20 = 0
a. Find the corresponding standard form
b. Identify the center and radius of the circle
c. Find the x- and y-intercepts
d. Graph the circle

Answer 1

no I am not sure how to even begin this problem

Answer 2

The standard form of the equation of any circumference has the form:
\[(x-\alpha)^2+(y-\beta)^2=r^2\]
Where \(C(\alpha, \beta)\) are the coordinates of the center and "r" is the radius of the circle in question.
Now there are several ways to turn the general form to the standard one, one of them is completing the squares.
Completing the squares relies solely on the structure: \((x-a)^2=x^2+2ax+a^2\) so you conveniently add or sustract terms in order to obtain the structure of a perfect square to bring it to the form \((x-\alpha)^2+(y-\beta)^2=r^2\).

Another way is to pull out the radius and the coordinates of the center out of the given equation, you know that a circles general equation has the form:
\[x^2+y^2+Dx+Ey+F=0\]
Where:
\[\alpha=-\frac{ D }{ 2 }\]
\[\beta = -\frac{ E }{ 2 }\]
\[r=\sqrt{\alpha ^2 + \beta^2-F}\]

Answer 3

\[x ^{2}+2*x*5+5^{2}-5^{2}+y ^{2}-2*y*2 +2^{2}-2^{2}+20=0\]

Answer 4

we are writing the given equation in the form of \[(a+b)^{2}= a ^{2}+2ab+b ^{2} and (a-b)^{2}= a ^{2}-2ab+b ^{2}\]

Answer 5

(\[(x+5)^{2}+(y-2)^{2}-25-4+20=0\]

Answer 6

\[(x+5)^{2}+(y-2)^{2}=9\]

Answer 7

\[[x-(-5)]^{2}+(y-2)^{2}=3^{2}\]

Answer 8

this is the standard form of the circle equation