Question

Write the standard equation of the circle with the given center that passes through the given point. Center (-2,6); point (-2,10)

Answer 1

The general form of the equation of a circle is given by:
(x-a)²+(y-b)²=r²
where:
(a,b) is the center
r is the radius
Given that the circle with center (-2,6) and cuts point (-2,10), the equation of the circle will be found as follows:
the radius of the circle will be:
r=√[(x-a)²+(y-b)²]
r=√[(-2-(-2))²+(10-6)²]
r=√[(-2+2)²+4²]
r=√[0+4²]
r=4 units 
hence plugging the values to obtain the equation we get:
(x-(-2))²+(y-6)²=4²
simplifying we get:
(x+2)²+(y-6)²=4²