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The endpoints of a diameter of a circle are (5,-10) and (-7,2). What is the equation of the circle in standard form?

User Clarybel
by
8.3k points

1 Answer

4 votes

Answer:


(x+1)^2+(y+4)^2=(6√(2))^2

Explanation:

The length of the diameter
(d) will be the distance between
(5,-10)\ and\ (-7,2).


d=√((2-(-10))^2+(-7-5)^2)=√((12)^2+(-12)^2)=√(144+144)=√(288)\\\\d=12√(2)

Radius:


radius(r)=(diameter)/(2)=(12√(2))/(2)=6√(2)

Centre:

Let
(a,b) be centre of the circle.

Centre will be the mid point of end points of diameter.


a=(5-7)/(2)=(-2)/(2)=-1\\\\b=(-10+2)/(2)=(-8)/(2)=-4\\\\Centre=(-1,-4)

Equation of circle:

If
(a,b) be centre and
r be the radius.

Equation of circle:
(x-a)^2+(y-b)^2=r^2

Here
(a,b)=(-1,-4)\ and\ r=6√(2)


(x-(-1))^2+(y-(4))^2=(6√(2))^2\\(x+1)^2+(y+4)^2=(6√(2))^2

User Syrkull
by
9.0k points

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