Question

write the standard form of the equation of the circle described below. center (0,2) passes through the point (1, - 4) the standard form of the equation of the circle is . (type an equation. simplify your answer.)

Explanation:

Step1: Recall circle - equation formula

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center of the circle and $r$ is the radius. Here, $h = 0$ and $k = 2$, so the equation is $(x - 0)^2+(y - 2)^2=r^2$, or $x^{2}+(y - 2)^2=r^2$.

Step2: Calculate the radius

The radius $r$ is the distance between the center $(0,2)$ and the point $(1,-4)$ on the circle. Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here, $x_1 = 0,y_1 = 2,x_2 = 1,y_2=-4$. Then $r=\sqrt{(1 - 0)^2+(-4 - 2)^2}=\sqrt{1+( - 6)^2}=\sqrt{1 + 36}=\sqrt{37}$.

Step3: Write the final - equation

Substitute $r=\sqrt{37}$ into the equation $x^{2}+(y - 2)^2=r^2$. We get $x^{2}+(y - 2)^2 = 37$.

Answer:

$x^{2}+(y - 2)^2 = 37$