I have this question:
Find the equations of the two circles which pass through the point $(2,0)$ and have both the $y$-axis and the line $y-1=0$ as tangents.
By plotting $y=1$ and $(2,0)$, it's obvious that the smaller circle has a radius of 1 and therefore an equation of $(x-1)^2 + y^2 = 1$. When you look at the larger one, it seems to touch these 3 co-ordinates: $(r,1)$, $(2,0)$, $(0, 1-r)$ where $r$ is its radius.
The larger circle's equation is apparently $(x-5)^2+(y+4)^2=25$. How do I go about finding this algebraically?
Best Answer
Let the equation of the circle be $(x-a)^2+(y-b)^2=r^2$ where $r\ge0$
Now the distance of any tangent from the centre of the circle = radius
For $y$ axis $x=0,r=\dfrac{|a|}{\sqrt{1^2+0^2}}=|a|\implies a=\pm r$
For $y-1=0, r=\dfrac{|b-1|}{\sqrt{1^2+0^2}}=|b-1|\implies r^2=(b-1)^2$
Case $\#1:$ If $b-1\ge0, r=b-1\iff b=r+1$
So, we have $(x\pm r)^2+\{y-(r+1)\}^2=r^2$
Now find $r(\ge0)$ using the fact : the circle passes through $(2,0)$
Case $\#2:$
If $b-1<0, r=-(b-1)=1-b\iff b=1-r$
Like $\#1$