Question

Q4 Help pleaseeeeeeee

Answer 1

a. The general equation for a circle centered at $(a,b)$ with radius $r$ is

$(x-a)^2+(y-b)^2=r^2$

The described circle has equation

$(x+3)^2+(y+2)^2=r^2$

We know the circle passes through the origin. This means that the equation above holds for $x=0$ and $y=0$. The distance between any point on the circle and its center is the radius, so we can use this fact to determine $r$:

$(0+3)^2+(0+2)^2=r^2\implies 9+4=13=r^2\implies r=\sqrt{13}$

So the circle's equation is

$(x+3)^2+(y+2)^2=(\sqrt{13})^2=13$

b. If the distance between point B and the center is less than $\sqrt{13}$, then B lies inside the circle. If the distance is greater than $\sqrt{13}$, it falls outside the circle. Otherwise, if the distance is exactly $\sqrt{13}$, then B lies on the circle.

The distance from B to the center is

$\sqrt{(-1+3)^2+(3+2)^2}=\sqrt{4+25}=\sqrt{29}$

$29>13$, so $\sqrt{29}>\sqrt{13}$, which means B falls outside the circle.