Question

$3$ circles of radii $a,b,c(a<b<c)$ touch each other externally and have $X-$ axis as a common tangent then

• A. $a,b,c$ are in A.P
• B. $\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}$
• C. $\sqrt{a},\sqrt{b},\sqrt{c}$ are in A.P
• D. $\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}$

Answer 1

The correct option is D

$\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}$

Explanation for correct option.

Step1. constructing diagram

Step2. Expressing the given data according to the question,

Given, circles of radii $a,b,c(a<b<c)$ touch each other externally and have $X-$ axis as a common tangent

Let,$O,B,A$ are the center of the circle

Then

$$\begin{gathered} \Rightarrow OQ=a \\ \Rightarrow BR=b \\ \Rightarrow AP=c \end{gathered}$$

Since, QP is perpendicular to AP & OE perpendicular to AP Therefore OQPE is rectangle & OE=PA

$$\begin{gathered} \Rightarrow OA=c+a \\ \Rightarrow AE=c-a \end{gathered}$$

Similarly,

$\begin{array}{rcl}& \Rightarrow & BF=b-a\\ & \Rightarrow & OB=b+a\end{array}$

Step3. Finding the value of PQ

Now, $△AOE$is a right angle triangle.

Therefore,

$\begin{array}{rcl}& \Rightarrow & {\left(OA\right)}^2={\left(AE\right)}^2+{\left(OE\right)}^2\\ & \Rightarrow & (OE{)}^2={\left(OA\right)}^2-{\left(AE\right)}^2\\ & \Rightarrow & {\left(OE\right)}^2={(c+a)}^2-{(c-a)}^2\\ & \Rightarrow & P{Q}^2={(c+a)}^2-{(c-a)}^2\\ & \Rightarrow & PQ=\sqrt{{(c+a)}^2-{(c-a)}^2}\\ & \Rightarrow & PQ=\sqrt{4ac}....\left(i\right)\end{array}$

Step4.Finding value of QR

$△BOF$ is right angle triangle

therefore,

$$\begin{gathered} \Rightarrow {\left(OB\right)}^2={\left(BF\right)}^2+{\left(OF\right)}^2 \\ \Rightarrow {\left(OF\right)}^2={\left(OB\right)}^2-{\left(BF\right)}^2 \\ \Rightarrow {\left(OF\right)}^2={(b+a)}^2-{(b-a)}^2 \\ \Rightarrow Q{R}^2={(b+a)}^2-{(b-a)}^2 \\ \Rightarrow QR=\sqrt{{(b+a)}^2-{(b-a)}^2} \\ \Rightarrow QR=\sqrt{4ab}....\left(ii\right) \end{gathered}$$

Step5. Finding value of PR

Draw BM perpendicular to AP

Then,

$$\begin{gathered} \Rightarrow AB=c+b \\ \Rightarrow AM=c-b \end{gathered}$$

Now,

$$\begin{gathered} \Rightarrow A{B}^2=B{M}^2+A{M}^2 \\ \Rightarrow B{M}^2=A{B}^2-A{M}^2 \\ \Rightarrow P{R}^2={(c+b)}^2-{(c-b)}^2 \\ \Rightarrow PR=\sqrt{{(c+b)}^2-{(c-b)}^2} \\ \Rightarrow PR=\sqrt{4bc}....\left(iii\right) \end{gathered}$$

Step6. Finding relation between radius

$\Rightarrow PR=PQ+RQ....\left(iv\right)$

From equation $\left(i\right),\left(ii\right),\left(iii\right)\&\left(iv\right)$

$\Rightarrow \sqrt{4bc}=\sqrt{4ab}+\sqrt{4ac}$

Divide with $\sqrt{4abc}$,

$\Rightarrow \frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}$

Hence, correct option is $\mathbf{\left(}\mathit{D}\mathbf{\right)}$.