find the equation of the circle which end points of the diameter is (2,-5) and (6,8)
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the equation of the circle is
(x-2)(x-6)+{y-(-5)}(y-8)=0
Or, (x-2)(x-6)+(y+5)(y-8)=0
Or, x^2-6x-2x+12+y^2-8y+5y-40=0
Or, x^2+y^2-8x-3y-28=0
(x-2)(x-6)+{y-(-5)}(y-8)=0
Or, (x-2)(x-6)+(y+5)(y-8)=0
Or, x^2-6x-2x+12+y^2-8y+5y-40=0
Or, x^2+y^2-8x-3y-28=0
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