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Find the general equation of the circle having a diameter with endpoints at A(-2,3) and B(4,5).

User Tzim
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Answer:
(x-1)^2+(y-4)^2=\text{ 10}Step-by-step explanation:

The equation of a circle whose center is (a, b) and which has a radius r is given as:

(x - a)² + (y - b)² = r²

For the circle with endpoints at A(-2,3) and B(4,5).​

The center (a, b) of the circle is calculated as:


\begin{gathered} a=(-2+4)/(2) \\ a=(2)/(2) \\ a=1 \\ b=(3+5)/(2) \\ b=(8)/(2) \\ b=4 \end{gathered}

The center, (a, b) = (1, 4)

The diameter(D) of the circle is the distance between the endpoints A(-2,3) and B(4,5).​


\begin{gathered} D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ D=\sqrt[]{(4-(-2))^2+(5-3)^2} \\ D=\sqrt[]{(4+2)^2+2^2} \\ D=\sqrt[]{6^2+2^2} \\ D=\sqrt[]{36+4} \\ D=\sqrt[]{40} \end{gathered}

The diameter of the circle = √40

Radius = Diameter / 2

r = d/2

r = √40 / 2

Substituting a = 1, b = 4, and r = √40 / 2 into the general equation of a circle:


\begin{gathered} (x-1)^2+(y-4)^2=(\frac{\sqrt[]{40}}{2})^2 \\ (x-1)^2+(y-4)^2=(40)/(4) \\ (x-1)^2+(y-4)^2=\text{ 10} \end{gathered}

The general equation of the circle is:


(x-1)^2+(y-4)^2=\text{ 10}

User Sajeer Babu
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