The equation of a circle whose diameter is the line connecting the points (1, 5) and (9, 1) is?

1 Answer
CW
Nov 19, 2017

(x-5)^2+(y-3)^2=20

Explanation:

Equation of a circle with center at(h,k) and a radius r is given by : (x-h)^2+(y-k)^2=r^2
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As AB is the diameter of the circle, the midpoint O of AB is the center of the circle,
=> midpoint O=((1+9)/2,(5+1)/2)=(5,3)
=> r^2=(9-5)^2+(1-3)^2=20
Hence, equation of the circle is :(x-5)^2+(y-3)^2=20

Solution 2) :
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Let P be any point on the circle,
as AB is the diameter, => angleAPB=90^@
let m_1, m_2 be the slopes of AP and BP, respectively,
=> m_1=(y-5)/(x-1),
=> m_2=(y-1)/(x-9),
since AP and BP are perpendicular to each other, their products of the slopes is -1, => m_1xxm_2=-1,
=> (y-5)/(x-1)*(y-1)/(x-9)=-1
=> (x-1)(x-9)=-(y-5)(y-1)
=> (x-1)(x-9)+(y-5)(y-1)=0
=> x^2-10x+y^2-6y=-14
=> (x-5)^2-25+(y-3)^2-9=-14
=> (x-5)^2+(y-3)^2=-14+25+9
=> (x-5)^2+(y-3)^2=20

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