Question

Find an equation of the circle that has center (-4,3) and passes through (6,-4).

Answer 1

Answer:

$(x+4)^2+(y-3)^2=149$

Step-by-step explanation:

The 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 of the circle.

Given that $(h,k)\rightarrow(-4,3)$ and it passes $(6,-4)$, their distance between each other must the radius of the circle, so we can use the distance formula to find the radius:

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\\\d=\sqrt{(-4-3)^2+(6-(-4))^2}\\\\d=\sqrt{(-7)^2+10^2}\\\\d=\sqrt{49+100}\\\\d=\sqrt{149}$

Therefore, if the length of the radius is $r=\sqrt{149}$ units, then $r^2=149$, making the final equation of the circle $(x+4)^2+(y-3)^2=149$