Question

find the standard form of the equation of the circle having the following properties. center at the origin containing the point (5, - 7) type the standard form of the equation of this circle.

Explanation:

Step1: Recall the standard - form of a circle equation

The standard - form of a circle equation is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center of the circle and $r$ is the radius. Here, the center of the circle is at the origin $(0,0)$, so $h = 0$ and $k = 0$. The circle contains the point $(5,-7)$.

Step2: Calculate the radius

The radius $r$ is the distance between the center $(0,0)$ and the point $(5,-7)$ on the circle. Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, we have $r=\sqrt{(5 - 0)^2+(-7 - 0)^2}=\sqrt{25 + 49}=\sqrt{74}$.

Step3: Write the circle equation

Substitute $h = 0$, $k = 0$, and $r=\sqrt{74}$ into the standard - form equation. We get $(x - 0)^2+(y - 0)^2=(\sqrt{74})^2$, which simplifies to $x^{2}+y^{2}=74$.

Answer:

$x^{2}+y^{2}=74$