Question

find the standard form of the equation of the circle with endpoints of a diameter at the points (5,8) and (-9,4). type the standard form of the equation of this circle. (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 $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here, $x_1 = 5,y_1 = 8,x_2=-9,y_2 = 4$. So the center $(h,k)=(\frac{5+( - 9)}{2},\frac{8 + 4}{2})=(\frac{5-9}{2},\frac{12}{2})=(-2,6)$.

Step2: Find the radius of the circle

The radius $r$ is the distance between the center $(h,k)=(-2,6)$ and one of the endpoints of the diameter, say $(x_1,y_1)=(5,8)$. The distance formula is $r=\sqrt{(x_1 - h)^2+(y_1 - k)^2}$. Substitute the values: $r=\sqrt{(5-( - 2))^2+(8 - 6)^2}=\sqrt{(5 + 2)^2+2^2}=\sqrt{49 + 4}=\sqrt{53}$.

Step3: Write the standard form of the circle equation

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$. Substitute $h=-2,k = 6,r=\sqrt{53}$ into the equation, we get $(x+2)^2+(y - 6)^2=53$.

Answer:

$(x + 2)^2+(y - 6)^2=53$