Find an equation of the circle whose diameter has endpoints (-5-4) and (3-6)

Question

Find an equation of the circle whose diameter has endpoints (−5−4) and (3−6)

Answer 1

First we must find the center of the circle, which is the midpoint of the diameter.

 

The midpoint of a line segment whose endpoints are (x₁,y₁) and (x_2,,y₂) is

 

M = ((x₁ + x₂₎)/2,(y₁ + y₂)/2).

 

So, the midpoint of the segment whose endpoints are (-5,-4) and (3,-6) is ((-5 + 3)/2,(-4 + -6)/2) = (-1,-5)

 

The point (-1,-5) is the center of the circle.

 

Next, we must find the length of the radius of the circle, which is the segment starting at the center and extending to a point on the circle.  The center is (-1,-5), and we know that (-5,-4) and (3,-6) are points on the circle.  I will show the calculation of the radius using each point on the circle, even though you only need to use one point.

 

If we use (-5,-4), the distance formula tells us that the distance from (-1,-5) to (-5,-4) is

 

√(-5 - (-1))² + (-4 - (-5))²  = √16 + 1 = √17

 

So of the radius of the circle is √17.

 

The  equation for a circle with center at (a,b) and radius r is

(x - a)² + (y - b)² = r²

 

Now we will substitute a = -1, b = -5, and r = √17

 

The equation of the circle is (x - (-1))² + (y - (-5))² = (√17)².

 

Simplifying, we (x + 1)² + (y + 5)² = 17

Answer 2

the equation of a circle is (x-h)^2+(y-k)^2=r^2 where (h,k) is the center and r is the length of the radius

first find the midpoint of the diameter which is the center of the circle

points (-5,-4) and (3,-6)

x=(-5+3)/2=-2/2=-1

y=(-4-6)/2=(-10)/2=-5

the center is (h,k)=(-1,-5)

now find the length of the radius using the center and either one of the endpoints

use the distance formula √[(x1-x2)^2+(y1-y2)^2]

using (-5,-4) and (-1,-5)...

√(-4)^2+(1)^2=√16+1=√17 which again is the radius

now substitute into the general formula

(x+1)^2+(y+5)^2=(√17)^2, or

(x+1)²+(y+5)²=17