equation in standard form

Question

The end points of a diameter of a circle are (-2,-3) and (-10,-9)

 

Write an equation of the circle in standard form.

Answer 1

diameter of a circle with endpoints (-2,-3) and (-10,-9)

form a right triangle with the diameter being the hypotenuse

one side is l(-9-[-3])l=l-6l=6 units (vertical side)

the other side is l(-10-[-2])l=l-8l=8 (horizontal side)

use the Pythagorean Theorem H^2=6^2+8^2

H^2=36+64

H^2=100

H=10 units for the diameter (or if you realized that you had a variation of a 3-4-5 right triangle then H=2*5=10)

the radius is 5 units

the midpoint of the diameter which is the center of the circle is (-9+[-3])/2=-12/2=-6 for k

(-10+[-2])/2=-12/2=-6 for h

the center is (h,k)=(-6,-6) and the radius is 5

(x-h)^2+(y-k)^2=r^2 is the equation of a circle

(x+6)^2+(y+6)^2=5^2

(x+6)²+(y+6)²=25

                                    

 

Answer 2

We need to first find the center (h,k) of the circle and then the radius squared r^2

 

In order to find the center of the circle we need to use the mid point equation of our line which is

 

[(x₁+x₂)/2 , (y₁+y₂) / 2]

 

= (-2-10)/ 2 , (-3-9)/2

 

= -6, -6

 

 

So this is our (h,k)

 

Next we need to find the radius of our circle which can be done by choosing an arbitrary point on the cirlce

 

let it be (-2,-3)

 

r^2 = (-2+6)^2  + (-3+6)^2 

 

r^2 = 16 + 9

 

r = 5

 

So our final answer would be :

 

(x+6)^2 + (y+6)^2 = 25