Question

What is the equation of the circle with center (4, 4) that passes through the point (10, 14)? HELP MEEEEE PLEASE

Answer 1

Answer:D

Step-by-step explanation:

Answer 2

Answer:

The equation of the circle can be written as:

• $\left(x-4\right)^2+\left(y-4\right)^2=136$

Step-by-step explanation:

The general equation of a circle with center $(h,k)$ and radius $r$ is:

$\left(x-h\right)^2+\left(y-k\right)^2=r^2$

In our example, we know $\left(h,k\right)=\left(4,4\right)$, as we just have to make sure we need determine $\:r^2$.

$\left(x-4\right)^2+\left(y-4\right)^2=r^2\:\:$

As the circle passes through (10, 14), that pair of values for x and y must satisfy the equation. So we have:

$\left(10-4\right)^2+\left(14-4\right)^2=r^2$

$\mathrm{Switch\:sides}$

$r^2=\left(10-4\right)^2+\left(14-4\right)^2$

$r^2=6^2+10^2$

$r^2=36+100$

$r^2=136$

Thus the equation of the circle can be written as:

$\left(x-4\right)^2+\left(y-4\right)^2=136$