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find the standard form of the equation of the circle with endpoints of …

Question

find the standard form of the equation of the circle with endpoints of a diameter at the points (5,4) and (-1,6). (type an equation.)

Explanation:

Step1: Find the center of the circle

The center of the circle is the mid - point of the diameter. The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Given $(x_1,y_1)=(5,4)$ and $(x_2,y_2)=(-1,6)$, the center $(h,k)=(\frac{5+( - 1)}{2},\frac{4 + 6}{2})=(2,5)$.

Step2: Find the radius of the circle

The radius $r$ is the distance from the center $(h,k)$ to either of the endpoints of the diameter. The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Using the center $(2,5)$ and the point $(5,4)$, we have $r=\sqrt{(5 - 2)^2+(4 - 5)^2}=\sqrt{9 + 1}=\sqrt{10}$.

Step3: Write the standard form of the circle equation

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$. Substituting $h = 2$, $k = 5$, and $r=\sqrt{10}$, we get $(x - 2)^2+(y - 5)^2 = 10$.

Answer:

$(x - 2)^2+(y - 5)^2 = 10$