214k views
3 votes
what is the general form of the equation of a circle with its Center at -2, 1 and passing through -4, 1

1 Answer

4 votes

The given center is (-2,1), which means h = -2 and k = 1. We also know that the circle passes through (-4,1), which means x = -4 and y = 1.

Let's use the standard form first


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

Let's replace the given information to find r


\begin{gathered} (-4-(-2))^2+(1-1)^2=r^2 \\ r=\sqrt[]{(-4+2)^2}=\sqrt[]{(-2)^2} \\ r=\sqrt[]{4}=\pm2 \end{gathered}

Once we have r, we can look for the general form


(x-(-2))^2+(y-1)^2=4

Then, we solve the binomials


\begin{gathered} (x+2)^2+(y-1)^2=4 \\ x^2+4x+4+y^2-2y+1=4 \end{gathered}

At last, we organize all the terms and the left side.


x^2+y^2+4x-2y+4+1-4=0

Hence, the general form is


x^2+y^2+4x-2y+1=0

User Luka Kama
by
7.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories