Question

write the standard equation for the circle. center at (7,3) and passing through the origin. choose the correct answer below. a. (x - 7)^2+(y - 3)^2 = \sqrt{58} b. (x - 7)^2+(y - 3)^2 = 58 c. (x + 7)^2+(y + 3)^2 = 58 d. (x + 7)^2+(y + 3)^2 = \sqrt{58} e. x^2 + y^2 = 58 f. x^2 + y^2 = 49 g. x^2 + y^2 = 9 h. x^2 + y^2 = \sqrt{58}

Explanation:

Step1: Recall circle - standard equation

The standard 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, the center is $(7,3)$, so the equation is $(x - 7)^2+(y - 3)^2=r^2$.

Step2: Calculate the radius

The circle passes through the origin $(0,0)$. The distance between the center $(7,3)$ and the point $(0,0)$ is the radius $r$. Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, we have $r=\sqrt{(7 - 0)^2+(3 - 0)^2}=\sqrt{49 + 9}=\sqrt{58}$, and $r^2 = 58$.

Step3: Write the final equation

Substitute $r^2 = 58$ into the equation $(x - 7)^2+(y - 3)^2=r^2$, we get $(x - 7)^2+(y - 3)^2=58$.

Answer:

B. $(x - 7)^2+(y - 3)^2=58$