Answer:
• (a) (x-4)²+(y-2)²=4
,• (b) x²+y²-8x-4y+16=0
Explanation:
The given circle has the endpoints of one diameter as (2,2) and (6,2).
[tex]\begin{gathered} \text{Diameter=6-2=4 units} \\ \text{Radius, }r=\frac{4}{2}=2\text{ units} \end{gathered}[/tex]The centre of the circle is the midpoint of the diameter.
[tex]\begin{gathered} \text{Midpoint}=(\frac{2+6}{2},\frac{2+2}{2})_{} \\ =(\frac{8}{2},\frac{4}{2}) \\ =(4,2) \end{gathered}[/tex]Part A
The center-radius form of the equation of a circle is given as:
[tex](x-a)^2+(y-b)^2=r^2[/tex]Substitute the center, (a,b)=(4,2) and r=2:
[tex]\begin{gathered} (x-4)^2+(y-2)^2=2^2 \\ \implies(x-4)^2+(y-2)^2=4 \end{gathered}[/tex]Part B
The general form of a circle is given as:
[tex]x^2+y^2+2gx+2fy+c=0[/tex]In this case, we obtain the equation by expanding our results from part (a).
[tex]\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}[/tex]The general form of the circle is given as:
[tex]x^2+y^2-8x-4y+16^{}=0[/tex]
