Question

2.2.23Use the graph below to determine the equation of the circle in (a) center-radius form and (b) general form.a. Type thA10-(Simplify)(4,4)(2.2)(6,2)-10(4,0)10-10

Answer 1

• (a) (x-4)²+(y-2)²=4

,

• (b) x²+y²-8x-4y+16=0

Step-by-step explanation:

The given circle has the endpoints of one diameter as (2,2) and (6,2).

`\begin{gathered} \text{Diameter=6-2=4 units} \\ \text{Radius, }r=(4)/(2)=2\text{ units} \end{gathered}`

The centre of the circle is the midpoint of the diameter.

`\begin{gathered} \text{Midpoint}=((2+6)/(2),(2+2)/(2))_{} \\ =((8)/(2),(4)/(2)) \\ =(4,2) \end{gathered}`

Part A

The center-radius form of the equation of a circle is given as:

`(x-a)^2+(y-b)^2=r^2`

Substitute the center, (a,b)=(4,2) and r=2:

`\begin{gathered} (x-4)^2+(y-2)^2=2^2 \\ \implies(x-4)^2+(y-2)^2=4 \end{gathered}`

Part B

The general form of a circle is given as:

`x^2+y^2+2gx+2fy+c=0`

In this case, we obtain the equation by expanding our results from part (a).

`\begin{gathered} (x-4)^2+(y-2)^2=4 \\ x^2-8x+16^{}+y^2-4y+4=4 \\ x^2+y^2-8x-4y+16^{}+4-4=0 \\ x^2+y^2-8x-4y+16^{}=0 \end{gathered}`

The general form of the circle is given as:

`x^2+y^2-8x-4y+16^{}=0`