# Circles of radius $2$ passing through origin with centers on $x=1$

There are two circles of radius $2$ that have centers on the line $x=1$ and pass through the origin. Find their equations.

Please explain to me what the problem is really saying.

Here is a small diagram representing question:

You have to find equations of circles.

There centres are $(1,\alpha) \ and (1,-\alpha)$ by symmetry .

And the radius is $2$ units.

let equation of circles are : $x^2+y^2-2x\pm2\alpha y=0$ [c=0; as circles pass through (0,0)] ; then $R=\sqrt{\alpha^2+1}=2$

• what does x=1 mean in the problem? Does it mean 1 unit of the x-axis? – Samama Fahim Mar 31 '13 at 15:24
• in the diagram, is any of the circle passing through the origin? – Samama Fahim Mar 31 '13 at 15:30
• @SamamaFahim x=1 is the equation of the vertical line . And both pass through origin.Origin is intersection of X-axis and Y-axis – ABC Mar 31 '13 at 15:37
• how did you know that x=1 is a vertical line? – Samama Fahim Mar 31 '13 at 15:46
• @SamamaFahim Okay,You need to study some basics of co-ordinate geometry. As equation is $x=1$ , does not depend on $y$ so it must be locus of those points which have x-coordinate =1 independent of y. So, it must be a vertical line (x=constant,y is varying) passing through (1,0). – ABC Mar 31 '13 at 15:49

A circle with centre $(a,b)$ and radius $r>0$ has equation $(x-a)^2+(y-b)^2=r^2$. Since these circles have centre on the line $x=1$ and radius $2$, they have equation $(x-1)^2+(y-b)^2=4$. Since the circles pass through the point $(0,0)$, we have $(0-1)^2+(0-b)^2=4$, so $b^2+1=4$. Solving for $b$ gives $b=\pm\sqrt 3$, so the circles are given by $(x-1)^2+(y-\sqrt 3)^2=4$ and $(x-1)^2+(y+\sqrt 3)^2=4$.

• Is this equation correct $(0−1)^2+(0−b)^2=4$ for both the circles? – Samama Fahim Mar 31 '13 at 15:34
• Do both the circles have same place on the plane, do they coincide? – Samama Fahim Mar 31 '13 at 15:38