Question

find the standard form of the equation of the circle with endpoints of a diameter at the points (9,4) and (-3,6). 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, $x_1=9,y_1 = 4,x_2=-3,y_2 = 6$. So the center $(h,k)=(\frac{9+( - 3)}{2},\frac{4 + 6}{2})=(3,5)$.

Step2: Find the radius of the circle

The radius $r$ is the distance between the center $(h,k)=(3,5)$ and one of the endpoints of the diameter, say $(9,4)$. The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. So $r=\sqrt{(9 - 3)^2+(4 - 5)^2}=\sqrt{36+1}=\sqrt{37}$.

Step3: Write the standard form of the circle equation

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$. Substituting $h = 3,k = 5,r=\sqrt{37}$, we get $(x - 3)^2+(y - 5)^2=37$.

Answer:

$(x - 3)^2+(y - 5)^2=37$