When does the origin lie inside this circle?

When does the origin lie inside this circle?

Question

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

1. $c > 0$

---

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

---

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

---

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

---

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

Suggestion

Can we write this equation in the form $(x-p)^2+(x-q)^2=r^2$?

Which inequality denotes the region that is the inside of this circle?

Solution

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).