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Length of Internal Common Tangent

Length of int...

Question

Length of internal (transverse) common tangent to two circles is $L_{i n t} = \sqrt{d^{2} - (r_{1} - r_{2})^{2}}$

Where d = distance between the centers of the two circles $r_{1} & r_{2}$ are radii of two circles

A

True

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B

False

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Solution

The correct option is B
False

Let center of two circles being $c_{1} & c_{2}$ and their radii $r_{1} & r_{2}$

Then,

Given,

$c_{1} c_{2} = d$

Let transverse tangent touches the circle at A and B. It intersects the line segment $c_{1} c_{2} a t P .$

Let length of transverse tangent $A B = L_{t r a n s}$

Right angled triangle $c_{1} A P$

$C_{1} P^{2} = {r_{2}}^{2} + A P^{2}$ - - - - - - (1)

Right angles triangle $C_{2} B P$

$C_{2} P^{2} = {r_{2}}^{2} + P B^{2}$ - - - - - - (2)

In $△ A C_{1} P & △ B C_{2} P$

$∠ C_{1} A P = ∠ C_{2} B P = 90^{\circ}$

$∠ A P c_{1} = ∠ B P c_{2} = o p p o s i t e a n g l e$

Since, two angles of a triangle are equal, third angle should also be equal. {sum of the angle of triangle = $180^{\circ}$

$△ A C_{1} P ≅ △ B C_{2} P$ {AAA similarity}

$\frac{A c_{1}}{B c_{2}} = \frac{A p}{B p} = \frac{c_{1} p}{c_{2} p}$

$\frac{r_{2}}{r_{1}} = \frac{L_{t r a n s} - B P}{B P} = \frac{d - C_{2} P}{C_{2} P}$ $\to$ ______ $\leftarrow \frac{r_{2}}{r_{1}} = \frac{L_{t r a n s} r_{1}}{r_{1} + r_{2}} \frac{r_{2}}{r_{1}} = \frac{d - C_{2} P}{C_{2} P} C_{2} P = \frac{d r_{1}}{r_{1} + r_{2}}$

Substituting the values of $B P & c_{2} P$ in equation (2)

$\frac{d^{2} {r_{1}}^{2}}{(r_{1} + r_{2})^{2}} = {r_{1}}^{2} + \frac{{L^{2}}_{t r a n s} {r_{1}}^{2}}{(r_{1} + r_{2})^{2}}$

$d^{2} = (r_{1} + r_{2})^{2} + L_{t r a n s}^{2}$

$L_{t r a n s}^{2} = d^{2} - (r_{1} + r_{2})^{2}$

$L_{t r a n s} = \sqrt{d^{2} - (r_{1} + r_{2})^{2}}$

So,given statment

Length of internal (transverse) common tangent to circles is

$L_{i n t} = \sqrt{d^{2} - (r_{1} - r_{2})^{2}}$ is NOT correct.

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