Find the equation of the circle which has the line segment from a(-4,3) to b(2,1) as diameter.

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Solution

The centre and radius are needed.
Centre :
The centre is the mid point of ab.
Centre = $(\frac{- 4 + 2}{2}, \frac{3 + 1}{2}) = (- 1, 2)$
Radius :
The radius r is the distance from the centre (-1,2) to (-4,3)
$= \sqrt{(- 3)^{2} + (1)^{2}} = \sqrt{10}$
Equation :
$(x - h)^{2} + (y - k)^{2} = r^{2}$
$(x + 1)^{2} + (y - 2)^{2} = 10$

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Find the equation of the circle which has the line segment from a(-4,3) to b(2,1) as diameter.

The centre and radius are needed.Centre :The centre is the mid point of ab.Centre = (−4+22,3+12)=(−1,2)Radius :The radius r is the distance from the centre (-1,2) to (-4,3)=√(−3)2+(1)2=√10Equation :(x−h)2+(y−k)2=r2(x+1)2+(y−2)2=10