Question

What is the equation of a circle with diameter AB that has endpoints A(0, 0) and B(8, 6)?

Answer 1

Answer:

A. $(x-4)^2+(y-3)^2=25$

Step-by-step explanation:

Radius:

The length of diameter$(d)$ is the distance between A and B.

$d=\sqrt{(8-0)^2+(6-0)^2}=\sqrt{64+36}=\sqrt{100}=10\\\\radius(r)=\frac{d}{2}=\frac{10}{2}=5$

Centre:

Since A and B are end points of the diameter, centre is the mid point of these two. Let $(x,y)$ be the centre of the circle.

$x=\frac{8+0}{2}=4\\\\y=\frac{6+0}{2}=3$

Centre is $(4,3)$

Equation of circle:

If $(a,b)$ is the centre of the circle and $r$ be the radius. Equation of circle is given by:

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

$Here\ (a,b)=(4,3)\ and\ r=5\\Equation:\ (x-4)^2+(y-3)^2=5^2\\(x-4)^2+(y-3)^2=25$