Question

Determine the equation of the circle graphed below. ( please help me )

Answer 1

Answer:

$Equation : \\(x+3)^2 + ( y -2)^2 = 36$

Step-by-step explanation:

For standard form the circle's equation we need the centre of the circle and the radius.

Step 1: Find the centre

If the centre is not given find the end points of the diameter

and then find the mid point.

Let the end points of the diameter be : ( - 3 , 8 ) and ( -3 , -4 )

The mid-point of the diameter is :

                        $Mid-point = (\frac{-3 + - 3}{2}, \frac{-4+8}{2}) = (-3, 2)$

Therefore, centre of the circle = ( -3 , 2 )

Step 2 : Find radius

$Radius = \frac{Diameter }{2}$

Diameter is the distance between the end points ( -3 , 8) and ( -3 , -4 )

That is ,

 $Diameter = \sqrt{(-3-(-3))^2 + ( -4 -8)^2}$

                 $= \sqrt{(-3 + 3)^2 + (-12)^2}\\\\=\sqrt{0 + 144}\\\\=12$

Therefore ,

       $Radius = \frac{12}{2} = 6$

Step 3 : Equation of the circle

Standard equation of the circle with centre ( h ,k )

and radius ,r is :

       $(x - h)^2+(y -k)^2 = r^2$

Therefore, the equation of the circle with centre ( -3, 2)

and radius = 6 is :

    $(x - (-3))^2 + (y - 2)^2 = 6^2\\\\(x + 3)^2 + (y - 2)^2 = 36$