Question

find the equation of the circle that has a diameter with endpoints located at (-5 -3) and (-11 -3)

Answer 1

Answer: Option C

The equation is:

$(x+8)^2 +(y+3)^2=9$

Step-by-step explanation:

First we must calculate the midpoint between the two given points.

Then the midpoint will be the radius of the circumference

The midpoint between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

In this case the points are:

(-5 -3) and (-11 -3)

The the center is:

$(\frac{-5-11}{2}, \frac{-3-3}{2})$

$(-8,\ -3)$

Then the equation is:

$(x+8)^2 +(y+3)^2=r^2$

To find r we substitute one of the points in the equation and solve for r

$(-5+8)^2 +(-3+3)^2=r^2$

$(3)^2 +0=r^2$

$r^2 =3^2$

$r =3$

Finally the equation is:

$(x+8)^2 +(y+3)^2=9$

Answer 2

Answer:

Option C

Step-by-step explanation:

The standard form of equation of circle is:

(x-h)^2+ (y-k)^2=r^2

As we only know two points on the circle which are the ends of diameter.

As we know

Radius=Diameter/2

We have to find the length of diameter using the distance formula first to calculate radius. So,

Diameter= √((-11-(+5))^2+(-3-(-3)^2 )

= √((-11+5)^2+(-3+3)^2 )

=√((-6)^2+(0)^2 )

= √36

=6

Now,  

Radius=6/2

=3

As the diameter passes through centre, so the mid-point of diameter will be centre of the circle:

Mid-point=((x_1+x_2)/2,(y_1+y_2)/2)

=((-5-11)/2,(-3-3)/2)

=((-16)/2,(-6)/2)

=(-8,-3)

Putting the values of radius and centre in standard form

(x-h)^2+ (y-k)^2=r^2

(x-(-8))^2+ (y-(-3))^2=3^2

(x+8)^2+ (y+3)^2=9

So the correct answer is option C ..