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Find the equation of circles determined by the following conditions. The ends of diameter are (-5, 3) and (7, 5). |
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Answer» SOLUTION :The ENDPOINTS of diameter of the circle are (-5, 3) and (7, 5). `THEREFORE` EQU. Of the circle is `(x-h)^2 + (y-k)^2 = a^2` `(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0` or, `(x+5)(x-7)+(y-3)(y-5)=0` or, `x^2-7x+5x-35+y^2-5y-3y+15=0` or, `x^2+y^2-2x-8y-20=0` |
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