# Write the equation of a circle with endpoints of the diameter at (2, -5) and (-4, 3)

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### 2 Answers

The general equation of a circle with center at (h,k) and radius equal to r is given by (x - h)^2 + (y - k)^2 = r^2.

Here, we have the endpoints of a diameter as (2, -5) and (-4 , 3)

The mid point between these is the center of the circle. Or the center is [(2 - 4)/2 , (-5 + 3)/2] = (-1 , -1)

The radius of the circle is the half the length of the diameter.

=> (1/2)*sqrt [(2 + 4)^2 + (3 + 5)^2]

=> (1/2)*sqrt [ 6^2 + 8^2]

=> (1/2)*sqrt 100

=> 5

The equation of the circle with center (-1, -1) and radius 5 is

(x + 1)^2 +(y + 1)^2 = 25

**The equation of the circle with endpoints of a diameter (2, -5) and (-4 , 3) is (x + 1)^2 +(y + 1)^2 = 25**

Since the endpoints of the diameter are given, we'll calculate it's length.

D = sqrt[(-4-2)^2 + (3+5)^2]

D = sqrt(36 + 64)

D = sqrt100

D = 10 units.

If the diameter is 10, then the radius is R = D/2 = 5 units

To write the equation of the circle, we need the coordinates of the center and the length of the radius. So far, we have the length of the radius.

Still, we need the coordinates of the center.

We know that the center of the circle lies in the middle of the diameter of the circle.

xC = (2-4)/2 = -1

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

The coordinates of the center of circle are C(-1;-1).

We'll write the equation of the circle whose center is C(-1;-1) and radius is R = 5 units.

(x - xC)^2 + (y - yC)^2 = R^2

(x + 1)^2 + (y + 1)^2 = 25

We'll expand the squares:

x^2 + 2x+ 1 + y^2 + 2y + 1 - 25 = 0

x^2 + y^2 + 2x + 2y - 23 = 0

**The general equation of the circle is: x^2 + y^2 + 2x + 2y - 23 = 0.**