Question

What is the value of "a" for the following circle in general form?
pic attached

Answer 1

Firstly you should use Completion Of Square Method here to find the values of h which is one of the coordinate of center..

So:

Solve for this one only:

\[x^2 + ax \implies x^2 + ax + (\frac{a}{2})^2 - (\frac{a}{2})^2\]

We have nothing to do with last term here so it will become:

\[x^2 + ax \implies (x + \frac{a}{2})^2\]

Or if you have any doubt then I will show you whole:

\[x^2 + y^2 + ab + by + c = 0 \implies x^2 + ax + (\frac{a}{2})^2 + y^2 + by + (\frac{b}{2})^2 - (\frac{a}{2})^2 - (\frac{b}{2})^2 = -c\]

\[\implies (x + \frac{a}{2})^2 + (y + \frac{b}{2})^2 = (\frac{a}{2})^2 + (\frac{b}{2})^2 - c\]

Compare it with standard equation of the circle:

\[(x-h)^2 (y-k)^2 = r^2\]

Here :

\[h = \frac{-a}{2}\]

Now look at the graph and find center from that...

Answer 2

From my eyes, I am finding it as : \(3, -1\)

So compare it with \(h,k\)

\[h = 3\]

But h is:
\[h = \frac{-a}{2}\]

Equate them:

\[\frac{-a}{2} = 3\]

Multiply by -2 both the sides:

\[a = -6\]

Answer 3

oh ok cool! thanks soooo much!

Answer 4

Robtobey can provide you smaller and efficient solution I guess..

Answer 5

The general form of a cirlcle is:\[(x-h)^2+(y-k)^2=r^2\]where (h,k) is the center and r the radius. After substitution, expanding and simplifying, the following is the result:\[x^2-6 x+y^2+2 y+10=25 \]By inspection, a, the coefficient of x is -6.