Question

Write the standard form of the equation of the circle with the given center and radius.Center (−2,−5), r=6

Answer 1

Given, center of the circle (-2,-5)

The radius is r=6

Now the form of the equation of circle is:

$(x-h)^2+(y-k)^2=r^2$

Thus,

$\begin{gathered} (x-(-2))^2+(y-(-5))^2=6^2 \\ \Rightarrow(x+2)^2+(y+5)^2=36 \\ \Rightarrow x^2+4+4x+y^2+25+10y=36 \\ \Rightarrow x^2+y^2+4x+10y+29=36 \\ \Rightarrow x^2+y^2+4x+10y-7=0 \end{gathered}$

The answer is

$x^2+y^2+4x+10y-7=0$