Review question

# When does the origin lie inside this circle? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7599

## Solution

The origin lies within the circle with equation $x^2+ax+y^2+by=c$ precisely when

1. $c > 0$

2. $a^2 + b^2 > c$

3. $a^2 + b^2 < c$

4. $a^2 + b^2 > 4c$

5. $a^2 + b^2 < 4c$.

We can write the equation of the circle as

$\left(x+\dfrac{a}{2}\right)^2 -\left(\dfrac{a}{2}\right)^2+\left(x+\dfrac{b}{2}\right)^2 -\left(\dfrac{b}{2}\right)^2 = c,$

or

$\left(x+\dfrac{a}{2}\right)^2 +\left(x+\dfrac{b}{2}\right)^2 = c + \left(\dfrac{a}{2}\right)^2 +\left(\dfrac{b}{2}\right)^2.$

Thus the region inside the circle is denoted by

$\left(x+\dfrac{a}{2}\right)^2 +\left(x+\dfrac{b}{2}\right)^2 < c + \left(\dfrac{a}{2}\right)^2 +\left(\dfrac{b}{2}\right)^2.$

Does $(0,0)$ satisfy this inequality? If and only if

$\left(0+\dfrac{a}{2}\right)^2 +\left(0+\dfrac{b}{2}\right)^2 < c + \left(\dfrac{a}{2}\right)^2 +\left(\dfrac{b}{2}\right)^2,$

which is true if and only if $0 < c$, and the answer is (a).