Question

A circle of radius unity touches the co-ordinate axes in the first quadrant, then the sum of all the radii of the circles touching this circle and co-ordinate axes is
(a) 1
(b) 6
(c) 12
(d) 25

Answer 1

Hint: To solve this question, we should know the concept of the Pythagoras Theorem, that is, the sum of the squares of the perpendicular side and the base side is equivalent to the square of the hypotenuse. Also, we have to consider all the possible circles which touch the circle of radius unity and both the coordinate axes.

Complete step-by-step answer:
In this question, we have to find the sum of the radii of the circles touching the given circles of radius unity and coordinate axis. We will first draw the figure. According to the question, there are only two possible circles, that is one of them is below the circle of unity radius and another is above the circle of unity radius.

According to the figure, we are asked to find the sum of PR and QT, which is radii of both the possible cases. Let us consider that the radius of the smaller and larger circles be r and R respectively. In triangle SOC, we will apply the Pythagoras theorem. So, we can write,
\[{{\left( SC \right)}^{2}}+{{\left( OC \right)}^{2}}={{\left( OS \right)}^{2}}\]
Now, we have been given that SC = SD = 1 unit. And we also know that SD = OC which implies OC = 1 unit. So, we can write,
\[{{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{2}}={{\left( OS \right)}^{2}}\]
\[{{\left( OS \right)}^{2}}=2\]
\[OS=\sqrt{2}....\left( i \right)\]
Now, from the figure, we can see that OS = OR + RS, where RS is that radius of the circle with a unity radius. So, we can write,
OS = OR + 1
And we can also say that OR = OP + PR where PR is the radius of the smallest circle that PR = r.
OS = OP + r + 1
Now in \[\Delta POA\], we will apply Pythagoras theorem. So, we get,
\[{{\left( PA \right)}^{2}}+{{\left( OA \right)}^{2}}={{\left( OP \right)}^{2}}\]
And we know that PA = PB = r. Also, we know that OA = PB. So, we can write,
\[{{r}^{2}}+{{r}^{2}}={{\left( OP \right)}^{2}}\]
\[OP=r\sqrt{2}\]
So, we can write OS = OP + r + 1 as,
\[OS=r\sqrt{2}+r+1\]
Now, from equation (i), we will put the value of OS. So, we will get,
\[\sqrt{2}=r\sqrt{2}+r+1\]
\[r\left( \sqrt{2}+1 \right)=\sqrt{2}-1\]
\[r=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}....\left( ii \right)\]
Now, we will apply Pythagoras Theorem in \[\Delta QOE\]. So, we can write,
\[{{\left( OE \right)}^{2}}+{{\left( QE \right)}^{2}}={{\left( OQ \right)}^{2}}\]
And we know that QE = QF = R and we also know that OE = QF. So, we can write,
\[{{R}^{2}}+{{R}^{2}}=O{{Q}^{2}}\]
\[OQ=R\sqrt{2}\]
And from the figure, we can see that, OQ = OS + ST + TQ where ST is the radius of the circle with unity radius and TQ is the radius of the largest circle. So, we can write,
\[R\sqrt{2}=OS+1+R\]
And we know that \[OS=\sqrt{2}\], so we get,
\[R\sqrt{2}=\sqrt{2}+1+R\]
\[R\left( \sqrt{2}-1 \right)=\sqrt{2}+1\]
\[R=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}....\left( iii \right)\]
Now, we are asked to find the sum of the radii. So, we will find the value of the sum of R and r.
\[R+r=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}+\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\]
Now, we will take the LCM. So, we will get,
\[R+r=\dfrac{{{\left( \sqrt{2+1} \right)}^{2}}+{{\left( \sqrt{2}-1 \right)}^{2}}}{\left( \sqrt{2}-1 \right)\left( \sqrt{2}+1 \right)}\]
\[R+r=\dfrac{2+1+2\sqrt{2}+2+1-2\sqrt{2}}{2-1}\]
\[R+r=\dfrac{6}{1}\]
\[R+r=6\]
Hence, option (b) is the right answer.

Note: In this question, one might think of making circles in the second and fourth quadrant which is wrong. Because we are asked to find the sum of the radii of the circle which touches the circle of the unity radius and the coordinate axes which the unity radius circle touches.