86.1k views
4 votes
12. [Ex 2K) Find the centre, C, and the radius, r, of the following equation of circle: x2 + y2 - 6x + 4y + 9 = 0

User Shenequa
by
8.7k points

1 Answer

6 votes

An equation in the form:


(x-a)^2+(y-b)^2=r^2

is the standard form for the equation of a circle with center (a,b) and radius r. Here we have:


x^2+y^2-6x+4y+9=0

Then, group the x and y terms separately and "move" the constant to the right side of the equation:


x^2-6x+y^2+4y=-9

Complete the square:


x^2-6x+9+y^2+4y+4=-9+9+4

Factor:


(x-3)^2+(y+2)^2=4

Express the right side as a square:


(x-3)^2+(y-(-2))^2=2^2

Therefore:

The center is: (3, - 2), the radius is 2

Answer:


\begin{gathered} \text{Center: (3,-2)} \\ \text{Radius: 2} \end{gathered}

User NrNazifi
by
8.6k 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