Question

2.2.25Use the graph below to determine the equation of the circle in (a) center-radius form and (b) general form.a.10-(Sir(-5,6)Q6-7,4%(-3,4(-5,2)-1010-10-

Answer 1

Given the points on a circle,

(-5,6),(-3,4),(-7,4),(-5,2)

The general equation is circle is given by,

`\begin{gathered} x^2+y^2+2gx+2fy+c=0\ldots\ldots.(1) \\ \text{radius (r)=}\sqrt[]{g^2+f^2-c} \\ \text{centre (O)=(-g,-f)} \\ \text{Centre radius form is } \\ (x-h)^2+(y-b)^2=r^2 \end{gathered}`

For point (x,y)=(-5,6) the equation of circle becomes,

`\begin{gathered} (-5)^2+(6)^2+2g(-5)+2f(6)+c=0 \\ 25+36-10g+12f+c=0 \\ -10g+12f+c=-61\ldots\ldots(2) \end{gathered}`

For point (x,y)=(-3,4) the equation of circle becomes,

`\begin{gathered} (-3)^2+(4)^2+2g(-3)+2f(4)+c=0 \\ 9+16-6g+8f+c=0 \\ -6g+8f+c=-25\ldots\text{.}\mathrm{}(3) \end{gathered}`

For point (x,y)=(-5,2) the equation of circle becomes,

`\begin{gathered} (-5)^2+(2)^2+2g(-5)+2f(2)+c=0 \\ -10g+4f+c=-29\ldots\text{.}\mathrm{}(4) \end{gathered}`

For point (x,y)=(-7,4) the equation of circle becomes,

`\begin{gathered} (-7)^2+(4)^2+2g(-7)+2f(4)+c=0 \\ -14g+8f+c=-65\ldots\text{.}(5) \end{gathered}`