Question

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

Answer 1

4

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=√(16)=4`