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Find the equation of a circle, the co-ordinates of the ends of whose diameter are `(-1,-3) and (2,5)` Now the equation of circle `(x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0` `rArr(x+1)(x-2)+(y+3)(y-5)=0` `rArr x^(2)-x-2+ y^(2)-2y-15=0` `rArr x^(2)+y^(2)-x-2y-17=0` We is the required equation. |
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Answer» Here `(x_(1),y_(1))=(-1,-3) and (x_(2),y_(2))=(2,5)` Now the equation of circle `(x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0` `rArr(x+1)(x-2)+(y+3)(y-5)=0` `rArr x^(2)-x-2+ y^(2)-2y-15=0` `rArr x^(2)+y^(2)-x-2y-17=0` which is the required equation. |
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