Question

A circle has a radius of \sqrt{131} units and is centered at (0,-4.3). Write the equation of this circle.

Answer 1

radius = √131

and centre is = (0,-4.3)

so the equation is

(x-0)^2 +(y + 4.3)^2 = (√131)^2

x^2 + y^2 +8.6y - 112.51 =0

Answer 2

Answer:

x^ 2+(y+4.3)^2=131

Step-by-step explanation:

The strategy

The equation of a circle with center  

(h,k)(\blueD h, \maroonD k)

(h,k)

and radius  

r\goldD r

r

is given below.

(x−h)2+(y−k)2=r2(x - \blueD h)^2 + (y - \maroonD k)^2 = \goldD r^2

(x−h)

2

+(y−k)

2

=r

2

We can find the equation of our circle by substituting the given center and radius in this equation.

Hint #2

2 / 3

Writing the equation of the circle

We are given that the center of this circle is  

C=(0,−4.3)C=(\blueD{0},\maroonD{-4.3})

C=(0,−4.3)

and its radius is  

131\goldD{\sqrt{131}}

131

​

. Therefore, the standard equation of this circle can be written as follows.

(x−0)2+(y−(−4.3))2=(131)2(x - \blueD {0})^2 + (y - \maroonD {(-4.3)})^2 = (\goldD {\sqrt{131}})^2

(x−0)

2

+(y−(−4.3))

2

=(

131

​

)

2

We can simplify this equation by removing double negative signs and evaluating the expression on the right hand side of the equation.

x2+(y+4.3)2=131x^2 + (y +{4.3})^2 = {131}

x

2

+(y+4.3)

2

=131