Question

Show that the equation represents a circle by rewriting it in standard form, and find the center and radius of the circle.

5x^2 + 5y^2 + 10x − y = 0

Answer 1

Standard form for a circle is x^2 + y^2 == r^2.

Answer 2

\[5x^2+10x+5y^2-y=0\]
i just reordered the terms
\[5(x^2+2x)+5(y^2-\frac{1}{5}y)=0\]
\[(x^2+2x)+(y^2-\frac{1}{5}y)=0\]
i divided by sides by 5
\[(x^2+2x+(\frac{2}{2})^2)+(y^2-\frac{1}{5}y+(\frac{1}{5 \cdot 2})^2)=(\frac{2}{2})^2+(\frac{1}{5 \cdot 2})^2\]
now we have
\[(x^2+2x+1^2)+(y^2-\frac{1}{5}y+(\frac{1}{10})^2)=1+\frac{1}{100}\]
\[(x+1)^2+(y-\frac{1}{10})^2=\frac{101}{100}\]

Answer 3

center is \[(-1,\frac{1}{10})\]
radius is \[\sqrt{\frac{101}{100}}=\frac{\sqrt{101}}{10}\]

Answer 4

x^2+y^2+2x-y/5=0
=> (x+1)^2 -1 +(y-1/10)^2 -1/100=0
=>(x+1)^2 +(y-1/10)^2=101/100

this standard form of eqn of this circle..where center is
(−1,1/100) and radius =sqrt(101)/10