Question

What is the equation of the circle with center (5,-4) and containing the points (5,-2)

Answer 1

Answer:

The equation of a circle is $(x-5)^2+(y+4)^2=4$ .

Step-by-step explanation:

-The equation of a circle is:

$(x-h)^2+(y-k)^2=r^2$ (where center is $(h,k)$, the point $(x_{1},y_{1})$ and the radius known as $r$).

-Use the center (5,-4) and the point (5,-2) for the equation:

$(5-5)^2+(-2+4)^2=r^2$

-Solve the equation:

$(5-5)^2+(-2+4)^2=r^2$

$0 + (-2)^2=r^2$

$4 = r^2$

$\sqrt{4} = \sqrt{r^2}$

$2 = r$

-After you found the radius, use the center (5, -4)  and radius 2 to get the equation of a circle:

$(x-5)^2+(y+4)^2=2^2$

-Then, after you have the equation, the radius needs to simplified by the exponent:

$(x-5)^2+(y+4)^2=4$