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What is the equation of the circle that has a diameter joining the points (3, 6) and...

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What is the equation of the circle that has a diameter joining the points (3, 6) and (8, 10)

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justaguide's profile pic

Posted (Answer #1)

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The circle has a diameter that joins the points (3, 6) and (8, 10). The center of the circle is the point that lies between (3, 6) and (8, 10). This is ((3+8)/2, (6+10)/2) = (11/2, 8)

The radius of the circle is half the distance between the given points.

r = (1/2)*sqrt((8 - 3)^2 + (10 - 6)^2)

The equation of a circle with center (h, k) and radius r is:

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

=> (x - 11/2)^2 + (y - 8)^2 = 41/4

=> 4x^2 + 121 - 44x + 4y^2 + 256 - 64y = 41

=> 4x^2 + 4y^2 - 44x - 64y + 336 = 0

The equation of the circle is 4x^2 + 4y^2 - 44x - 64y + 336 = 0

aruv's profile pic

Posted (Answer #2)

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End points of diameter are (3,6) and (8,10).

Since angle in semi circle is right angle.Let (x,y) be any point in semicircle.

Find slope of lines joing end points of diameter.

`m_1=(y-6)/(x-3)`

`m_2=(y-10)/(x-8)`

`m_1 m_2={(y-6)/(x-3)}{(y-10)/(x-8)}=-1`

Equation of circle is

`(x-3)(x-8)+(y-6)(y-10)=0`

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