The point (-3,5) is on a circle with a center of ( 2,-6). Write the standard equation.

Answer 1

The standard 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.

first you will need to find the distance between the two given points-the center point and the point of the circle. That will give you the length of the radius. You can find that by plugging the two given points into the distance formula (Instead of d for distance I will use r to represent radius) (sqrt means square root....I don’t know how to do that symbol on here)

r=sqrt[(y2-y1)^2 + (x2-x1)^2] so

r=sqrt[(-6-5)^2 + (2-(-3))^2]

r=sqrt[(-11)^2 + (5)^2]

r=sqrt[121+25]

r=sqrt[146]

now that we have our radius, sqrt(146) and our center (2,-6) we can plug those into the formula.

(x-2)^2 + (y-(-6))^2 = [sqrt(146)]^2 which simplifies to

(x-2)^2 + (y+6)^2 = 146

Answer 2

TYPO: I did the final calculation wrong in my head somehow. The final answer should be 146, not 136!!! Otherwise, there you go!