Question 593105
Since ")" is "lower case" 0, I'm going to assume that point A is (2, 0).<br>
AP represents the length of segment AP (which is the distance between point A and point P. Using (x, y) for the point P and the distance formula  we get:
AP = {{{sqrt((x-2)^2 + (y-0)^2)}}}
which simplifies to:
AP = {{{sqrt(x^2-4x+4 + y^2)}}}<br>
Using the distance formula in a similar way we can find PB:
PB = {{{sqrt((x-(-1))^2 + (y-0)^2)}}}
Simplifying this we get:
PB = {{{sqrt((x+1)^2 + (y-0)^2)}}}
PB = {{{sqrt(x^2+2x+1 + y^2)}}}
<br>
Substituting these expressions into the given equation AP/PB = 2 we get:
{{{sqrt(x^2-4x+4 + y^2)/sqrt(x^2+2x+1 + y^2) = 2}}}<br>
Next we will transform this equation into the general form for conic sections:
{{{Ax^2 + Bxy + Cy^2 +Dx + Ey + F = 0}}}
We start by squaring both sides to eliminate the square roots:
{{{(x^2-4x+4 + y^2)/(x^2+2x+1 + y^2) = 4}}}
Next we will multiply each side by the denominator (to get rid of the fraction):
{{{((x^2-4x+4 + y^2)/(x^2+2x+1 + y^2))*(x^2+2x+1 + y^2) = 4*(x^2+2x+1 + y^2)}}}
which gives us:
{{{x^2-4x+4 + y^2 = 4x^2+8x+4 + 4y^2}}}
Now we just subtract {{{x^2-4x+4 + y^2}}} from each side. (The general form is best written with a positive leading coefficient.) This gives us:
{{{0 = 3x^2+12x + 3y^2}}}
Dividing each side by 3  we get:
{{{0 = x^2+ 4x + y^2}}}<br>
With A = C one would think that this would be a circle. But let's complete the squares to make sure:
{{{4 = (x+2)^2 + (y-0)^2}}}
We can now see that the equation is indeed the equation of a circle (with a center at (-2, 0) and a radius of 2).<br>
We needed to complete the squares because if we had gotten:<ul><li>{{{0 = (x+2)^2 + (y-0)^2}}}
we would have a "circle of radius 0" which is just a single point: (-2, 0).</li><li>any negative number = {{{(x+2)^2 + (y-0)^2}}}
we get nothing. This equation has NO solutions.</li></ul>