Question

Write the equation of the circle with center (-4, 8) and passes through the point (-2, -1)

Answer 1

Answer:

$(x+4)^2+(y-8)^2=85$

Step-by-step explanation:

The standard equation for a circle is given by:

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

Where (h, k) is the center and r is the radius.

We know that the center is (-4, 8). So, substitute -4 for h and 8 for k:

$(x-(-4))^2+(y-8)^2=r^2$

Simplify:

$(x+4)^2+(y-8)^2=r^2$

Now, we will need to find r.

We know that it passes through the point (-2, -1). So, we can substitute -2 for x and -1 for y and solve for r. So:

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

Evaluate:

$(2)^2+(-9)^2=r^2$

Square:

$4+81=r^2$

Add:

$r^2=85$

So, r squared is 85.

We don’t actually have to solve for r itself, since we will have to square it anyways.

So, we have:

$(x+4)^2+(y-8)^2=r^2$

Substituting 85 for r squared, we get:

$(x+4)^2+(y-8)^2=85$

And we have our equation.