find the equation of circle with ends of diameter at origin and (1,1)
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the answer is
x sqaure+y sqaure- 2x-5y+3=0
x sqaure+y sqaure- 2x-5y+3=0
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Solution
ends of diameter are (0,0,)=(x1,y1)
and (1,1)=(x2,y2)
now,
equation of the circle is
or, (x-x1)(x-x2)+(y-y1)(y-y2)=0
or, (x-0)(x-1)+(y-0)(y-1)=0
or, x(x-1)+y(y-1)=0
or, x^2-x+y^2-y=0
or, x^2+y^2-x-y=0
therefore the required equation is x^2+y^2-x-y=0
ends of diameter are (0,0,)=(x1,y1)
and (1,1)=(x2,y2)
now,
equation of the circle is
or, (x-x1)(x-x2)+(y-y1)(y-y2)=0
or, (x-0)(x-1)+(y-0)(y-1)=0
or, x(x-1)+y(y-1)=0
or, x^2-x+y^2-y=0
or, x^2+y^2-x-y=0
therefore the required equation is x^2+y^2-x-y=0
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