Question

2.write the equation of a circle with the following parameters Center at (0,-1)Passing through (-35,0)

Answer 1

Given:

`\begin{gathered} center\text{ }=(0,-1) \\ Through\text{ p}oint\text{ }(-35,0) \end{gathered}`

The equation of a circle is gotten by;

`\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ where: \\ x=-35 \\ y=0 \\ h=0 \\ k=-1 \\ \end{gathered}`

Substituting these values into the equation to get the value of r;

`\begin{gathered} (-35-0)^2+(0-(-1))^2=r^2 \\ (-35)^2+(1)^2=r^2 \\ 1225+1=r^2 \\ r^2=1226 \end{gathered}`

Thus, the equation of the circle is;

`\begin{gathered} (x-0)^2+(y-(-1))^2=1226 \\ x^2+(y+1)^2=1226 \end{gathered}`