Question

A circle has a diameter with endpoints at (6, 5) and (8, 5). Write the equation for the circle.

Answer 1

Answer:

$(x-7)^2+(y-5)^2=1$

Step-by-step explanation:

The two things that are required to formulate the equation of the circle is the center coordinate and the radius of the circle!

Center of the circle:

• The center of the circle always lies at the midpoint of the endpoints of its diameter: Let's call the endpoints A(6,5) and B(8,5).

Using the midpoint formula we'll get:

$(x_m, y_m) = \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$(x_m, y_m) = \left(\dfrac{6+8}{2},\dfrac{5+5}{2}\right)$

$(x_m, y_m) = (7,5)$

This is the center coordinate of our circle.

Radius:

The radius of the circle is the distance from the center of the circle to any of the endpoints of the diameter (A or B)

We can use the distance formula:

$r = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$r = \sqrt{(x_1-x_m)^2+(y_1-y_m)^2}$

$r = \sqrt{(6-7)^2+(5-5)^2}$

$r = \sqrt{1^2}$

$r = 1$

Equation of the circle:

The equation is written as:

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

here, (a,b) are the center points of the circle

in our case this is $(a,b)=(x_m,y_m)=(7,5)$

and r = 1

$(x-7)^2+(y-5)^2=1^2$

$(x-7)^2+(y-5)^2=1$

This is the equation of the circle!