Question

which equation represents a circle that contains the point (-2, 8) and has a center at (4, 0)? distance formula: $sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ $(x - 4)^2+y^2 = 100$ $(x - 4)^2+y^2 = 10$ $x^2+(y - 4)^2 = 10$ $x^2+(y - 4)^2 = 100$

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 $(h,k)=(4,0)$. So the equation of the circle is $(x - 4)^2+(y - 0)^2=r^2$, or $(x - 4)^2+y^2=r^2$.

Step2: Find the radius using the distance formula

The distance between the center $(4,0)$ and the point on the circle $(-2,8)$ is the radius $r$. Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, let $(x_1,y_1)=(4,0)$ and $(x_2,y_2)=(-2,8)$. Then $r=\sqrt{(-2 - 4)^2+(8 - 0)^2}=\sqrt{(-6)^2+8^2}=\sqrt{36 + 64}=\sqrt{100}=10$.

Step3: Substitute the radius into the circle equation

Since $r = 10$, then $r^2=100$. The equation of the circle is $(x - 4)^2+y^2=100$.

Answer:

$(x - 4)^2+y^2=100$