Question

find the standard form of the equation of the circle with endpoints of a diameter at the points (7,4) and (-5,2).
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, $x_1=7,y_1 = 4,x_2=-5,y_2 = 2$. So the center $(h,k)=(\frac{7+( - 5)}{2},\frac{4 + 2}{2})=(1,3)$.

Step2: Find the radius of the circle

The radius $r$ is the distance between the center $(h,k)=(1,3)$ and one of the endpoints of the diameter, say $(x_1,y_1)=(7,4)$. The distance formula is $r=\sqrt{(x_1 - h)^2+(y_1 - k)^2}$. Substitute the values: $r=\sqrt{(7 - 1)^2+(4 - 3)^2}=\sqrt{6^2+1^2}=\sqrt{36 + 1}=\sqrt{37}$.

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 $h = 1,k = 3,r=\sqrt{37}$ into the equation: $(x - 1)^2+(y - 3)^2=37$.

Answer:

$(x - 1)^2+(y - 3)^2=37$