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What is the equation of the circle whose diameter has endpoints (8,-2) and (-2,6)?

User NoobVB
by
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1 Answer

6 votes

Answer:


(x-3)^(2) +(y-2)^(2)=41

Explanation:

step 1

Find the diameter of the circle

the formula to calculate the distance between two points is equal to


d=\sqrt{(y2-y1)^(2)+(x2-x1)^(2)}

we have


A(8,-2)\\E(-2,6)

substitute the values


d=\sqrt{(6+2)^(2)+(-2-8)^(2)}


d=\sqrt{(8)^(2)+(-10)^(2)}


d=√(164)\ units


d=2√(41)\ units

step 2

Find the center of the circle

The center is the midpoint of the diameter

The center is equal to


C=((8-2)/(2),(-2+6)/(2))


C=(3,2)

step 3

Find the equation of the circle

The equation of the circle in center radius form is equal to


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

we have

(h,k)=(3,2)


r=2√(41)/2=√(41)\ units ---> the radius is half the diameter

substitute


(x-3)^(2) +(y-2)^(2)=(√(41))^(2)


(x-3)^(2) +(y-2)^(2)=41

User Diego Avila
by
8.4k points

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