Question

Write an equation for a circle with a diameter that has endpoints at (-10, 1) and (-8, 5). Round to the nearest tenth if necessary.

Answer 1

Answer:

The equation is (x+9)^2 + (y-3)^2 = 5

Step-by-step explanation:

The standard form for the equation of a circle is:

(x−h)^2+(y−k)^2=r2

The center of the circle is the midpoint of the diameter.

So the midpoint of the diameter at  (-10, 1) and (-8, 5) can be determined as:

(-10 +(-8))/2 , (1+5)/2

= -10-8/2, 1+5/2

= -18/2 , 6/2

= -9 , 3

Thus(-9,3) is the center of the circle.

Now we will use the distance formula to find the radius of the circle.

r^2=(-10-(-9))^2 + (1-3)^2

r^2=(-10+9)^2 +(-2)^2

r^2=(-1)^2 + (-2)^2

r^2=1 + 4

r^2= 5

Take square root at both sides.

√r^2= √5

r=√5

Now put the values in the 1st equation.

(x−h)^2+(y−k)^2=r2

where h = -9, k =3 and r = √5

(x-(-9))^2 + (y-3)^2= (√5)^2

(x+9)^2 + (y-3)^2 = 5

Thus the equation is (x+9)^2 + (y-3)^2 = 5 ....