Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Say I am asked to find, in expanded form without brackets, the equation of a circle with radius 6 and centre 2,3 - how would I go on about doing this?

I know the equation of a circle is $x^2 + y^2 = r^2$, but what do i do with this information?

share|improve this question
1  
$(x-a)^2 + (y-b)^2 = r^2$ is the general equation for the circle. Hopefully you can recognize what you have to do. –  Chinny84 May 29 at 13:15
    
The search phrase "equation of circle" gives a huge number of hits on Google, many of which immediately answer this question. –  user61527 May 30 at 0:01

4 Answers 4

The equation of a circle with centre $(a,b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$.

share|improve this answer
    
didn't see your answer pop up :). –  Chinny84 May 29 at 13:16

The equation of a circle with the centre at $(0,0)$ is $x^2+y^2=r^2$. This is because the circle with the radius $r$ is composed of all points which are $r$ away from $(0,0)$, and since the distance of a point $(x,y)$ from $(0,0)$ is $\sqrt{x^2+y^2}$, this means that the equation will be $$\sqrt{x^2+y^2}=r,$$ or, squaring that, $$x^2+y^2=r^2.$$

Hint

To center the point around an arbitrary point, think about how you would calculate the distance between $(x,y)$ and that arbitrary point.

share|improve this answer

The equation gets like this: $$(x-2)^2+(y-3)^2=36,$$which can bee seen as translating a circle with radius 6 and center $(0,0)$ (the equation you mentioned) to the new center $(2,3)$.

share|improve this answer

The general solution for the circle with centre $(a,b)$ and radius $r$ is $$ (x-a)^2+(y-b)^2=r^2. $$ Now, we have the centre $(2,3)$ and the radius $6$, therefore the equation of the circle is \begin{align} (x-2)^2+(y-3)^2&=6^2\\ x^2-4x+4+y^2-6x+9&=36\\ x^2+y^2-4x-6y+4+9-36&=0\\ \large\color{blue}{x^2+y^2-4x-6y-23}&\large\color{blue}{=0}. \end{align}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.