Question

Find an equation of the circle that has center (-2, 6) and passes through (-6, 1).

Answer 1

Answer:

(x+2)^2 + (y-6)^2 = 41

Step-by-step explanation:

The equation for a circle is (x-h)^2+(y-k)^2=r^2

So first, it is known that the circle's center is at (-2,6), Therefore, this can be filled in:

(x+2)^2+(y-6)^2=r^2

Next, we need to find the radius, and one of the points is already known, being (-6, 1)

With this, find the distance between these two points by doing the Pythagorean Theorem, a^2+b^2=c^2. The a^2 would be the x value changed and the b^2 would be the y value changed between the two numbers. Note that this is interchangeable.

To find a:

-2 to -6 = change of 4

To find b:

6 to 1 = change of 5

Next, write out the equation for this:

4^2+5^2=c^2

16+25=c^2

41=c^2

c = √41

The radius would be √41, so the equation can now be completed. Since c will be brought to the second power, this will cancel out the square root.

(x+2)^2 + (y-6)^2 = 41

Hope that helps.