Question

Each of the circles $x^2+y^2-2x-2y+1=0$ and $x^2+y^2+2x-2y+1=0$ touches internally a circle of radius $2$. The equation of circles touching all the three circles, is

• A. $x^2+{(y-\frac{7}{3})}^2=\frac{4}{9}$
• B. $x^2+{(y+\frac{1}{3})}^2=\frac{4}{9}$
• C. $x^2+{(y+\frac{7}{3})}^2=\frac{4}{9}$
• D. $x^2+{(y-\frac{1}{3})}^2=\frac{4}{9}$

Answer 1

Answer: (A), (B)

$x^2+{(y+\frac{1}{3})}^2=\frac{4}{9}$

$x^2+y^2-2x-2y+1=0\Rightarrow (x-1{)}^2+(y-1{)}^2=1$
and $x^2+y^2+2x-2y+1=0$
$\Rightarrow (x+1{)}^2+(y-1{)}^2=1$
Let the radius of the required circles be $r_1$ and centres at $(0,a)$ and $(0,b)$.
So, the possible figure of the required circles is


From the figure,
$\sqrt{(1+r_1{)}^2-1}=\sqrt{(a-1{)}^2}=a-1\cdots \left(1\right)[\because a>1]$
and $2=r_1+(a-1)\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$\sqrt{r_1^2+2r_1}=2-r_1\Rightarrow r_1^2+2r_1=4+r_1^2-4r_1\Rightarrow r_1=\frac{2}{3}$
From $\left(2\right)$, we get $a=\frac{7}{3}$
Hence, equation of the circle is $(x-0{)}^2+{(y-\frac{7}{3})}^2={\left(\frac{2}{3}\right)}^2$

Now, for the other circle with centre $(0,b)$
Since $b<1$, we have
$2=r_1+(1-b)\Rightarrow b=-\frac{1}{3}$
Hence, equation of the circle is $(x-0{)}^2+{(y+\frac{1}{3})}^2={\left(\frac{2}{3}\right)}^2$