Question

Standard equation of circle with center of (6,8) that passes through point (-1,4)

Answer 1

Answer:

The result should be:

$(x-6)^2+(y-8)^2=\sqrt{65}$

Step-by-step explanation:

-The equation of a circle:

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

-Use the center (6,8) and the point (-1,4) for this equation:

$(-1-6)^2+(4-8)^2=r^2$

-Then, you solve for the radius and get the equation:

$(-1-6)^2+(4-8)^2=r^2$

$(-7)^2+(-4)^2 =r^2$

$49 + 16 = r^2$

$65 = r^2$

$\sqrt{65} =\sqrt{r^2}$

$\sqrt{65} = r$

-Result:

$(x-6)^2+(y-8)^2=\sqrt{65}$