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Give the equation of a circle with a diameter that has endpoints (-7, 4) and (9,7).

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It is given that the endpoints of the diameter of the circle are (-7,4) and (9,7)

The diameter form is given by:


(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0

Therefore it follows:


\begin{gathered} (x-(-7))(x-9)+(y-4)(y-7)=0 \\ (x+7)(x-9)+(y-4)(y-7)=0 \\ x^2+7x-9x-63+y^2-4y-7y+28=0 \\ x^2+y^2-2x-11y-35=0 \end{gathered}

Hence the equation of the circle is given by:


x^2+y^2-2x-11y-35=0

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