Question

The radius of the circle whose equation is (x-3)^2 + (y+1)^2 = 16 is

Answer 1

Answer:

4

Step-by-step explanation:

The general equation of a circle with center (a, b) and radius r is given by the equation;

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

The constant in the right hand side of the equation is simply the square of the radius;

We have been given the following equation;

(x-3)^2 + (y+1)^2 = 16

Comparing this with the general equation above;

$r^{2}=16\\\\r=\sqrt{16}=4$