Name: Tangents to a Circle From an External Point
Uploaded: 2019-07-19
Duration: 3 min 5 s
Description: Do you know the equation of the tangents to a circle drawn from an external point? Let's find the equation involving the slope, which helps to find the equation of tangents.

Question

Let circles

S_{1}

and

S_{2}

of radii

r_{1}

and

r_{2}

respectively

(r_{1} > r_{2})

touches each other externally. Circle

S

of radius

r

touches

S_{1}

and

S_{2}

externally and also their direct common tangent. Prove that the triangle formed by joining centre of

S_{1}, S_{2}

and

S

is obtuse angled triangle.

Accepted Answer

Let, $A & B$ are the centres of two circles $S_{1} & S_{2}$ whose radii $r_{1} & r_{2}$ respectively touch each other externally.

Let, $C$ be the centre and $r$ be the radius of the circle $S$ which touches $S_{1}, S_{2}$ externally.

We know that if $O_{1} & O_{2}$ be the centres and $r_{1} & r_{2}$ be the radii of the circles $C_{1} & C_{2}$ respectively, then the length of a direct common tangent the circles $C_{1} & C_{2}$ is equal to $\sqrt{{(O_{1} O_{2})}^{2} - {(r_{1} - r_{2})}^{2}}$ .

Here, $D F, D E$ and $E F$ are the length of a direct common tangent the circles $S_{1} & S_{2}, S_{1} & S$ and $S & S_{2}$ respectively.

$\Rightarrow D F = D E + E F$

$\Rightarrow \sqrt{{(r_{1} + r_{2})}^{2} - {(r_{1} - r_{2})}^{2}} = \sqrt{{(r_{1} + r)}^{2} - {(r_{1} - r)}^{2}} + \sqrt{{(r + r_{2})}^{2} - {(r - r_{2})}^{2}}$

$\Rightarrow \sqrt{4 r_{1} r_{2}} = \sqrt{4 r_{1} r} + \sqrt{4 r r_{2}}$

$\Rightarrow {(\sqrt{r_{1} r_{2}})}^{2} = {(\sqrt{r_{1} r} + \sqrt{r r_{2}})}^{2}$

$\Rightarrow r_{1} r_{2} = r_{1} r + r_{2} r + 2 r \sqrt{r_{1} r_{2}}$

$\Rightarrow r = \frac{r_{1} r_{2}}{r_{1} + r_{2} + 2 \sqrt{r_{1} r_{2}}} . . . . (i)$

Apply the cosine rule in the triangle $△ A B C$

$\Rightarrow \cos (∠ A C B) = \frac{{(C A)}^{2} + {(C B)}^{2} - {(A B)}^{2}}{2 (C A) (C B)}$

$\Rightarrow \cos (∠ A C B) = \frac{{(r + r_{1})}^{2} + {(r + r_{2})}^{2} - {(r_{1} + r_{2})}^{2}}{2 (r + r_{1}) (r + r_{2})}$

$\Rightarrow \cos (∠ A C B) = \frac{r^{2} + {(r_{1})}^{2} + 2 r r_{1} + r^{2} + {(r_{2})}^{2} + 2 r r_{2} - ({(r_{1})}^{2} + {(r_{2})}^{2} + 2 r_{1} r_{2})}{2 (r^{2} + r (r_{1} + r_{2}) + r_{1} r_{2})}$

$\Rightarrow \cos (∠ A C B) = \frac{2 r^{2} + 2 r r_{1} + 2 r r_{2} - 2 r_{1} r_{2}}{2 (r^{2} + r (r_{1} + r_{2}) + r_{1} r_{2})}$

$\Rightarrow \cos (∠ A C B) = 1 - \frac{2 r_{1} r_{2}}{r^{2} + r (r_{1} + r_{2}) + r_{1} r_{2}}$

$\Rightarrow \cos (∠ A C B) = 1 - \frac{2 r_{1} r_{2}}{(r + r_{1}) (r + r_{2})}$

$\Rightarrow \cos (∠ A C B) = 1 - \frac{2 r_{1} r_{2}}{(\frac{r_{1} r_{2}}{r_{1} + r_{2} + 2 \sqrt{r_{1} r_{2}}} + r_{1}) (\frac{r_{1} r_{2}}{r_{1} + r_{2} + 2 \sqrt{r_{1} r_{2}}} + r_{2})}$

$\Rightarrow \cos (∠ A C B) = 1 - \frac{2}{(\frac{r_{2}}{r_{1} + r_{2} + 2 \sqrt{r_{1} r_{2}}} + 1) (\frac{r_{1}}{r_{1} + r_{2} + 2 \sqrt{r_{1} r_{2}}} + 1)}$

$\Rightarrow \cos (∠ A C B) = 1 - \frac{2 {(r_{1} + r_{2} + 2 \sqrt{r_{1} r_{2}})}^{2}}{(r_{1} + 2 r_{2} + 2 \sqrt{r_{1} r_{2}}) (2 r_{1} + r_{2} + 2 \sqrt{r_{1} r_{2}})}$

$\Rightarrow \frac{2 {(r_{1} + r_{2} + 2 \sqrt{r_{1} r_{2}})}^{2}}{(r_{1} + 2 r_{2} + 2 \sqrt{r_{1} r_{2}}) (2 r_{1} + r_{2} + 2 \sqrt{r_{1} r_{2}})} = 1 - \cos (∠ A C B) . . . . (i i)$

$\Rightarrow \frac{2 {(p + 1 + 2 \sqrt{p})}^{2}}{(p + 2 + 2 \sqrt{p}) (2 p + 1 + 2 \sqrt{p})} = 1 - \cos (∠ A C B) . . . . (i i),$ where $p = \frac{r_{1}}{r_{2}}$

Consider, $A = \frac{2 {(p + 1 + 2 \sqrt{p})}^{2}}{(p + 2 + 2 \sqrt{p}) (2 p + 1 + 2 \sqrt{p})}$

$\Rightarrow A = \frac{(2 p^{2} + 8 p \sqrt{p} + 12 p + 8 \sqrt{p} + 2)}{(2 p^{2} + 6 p \sqrt{p} + 5 p + 6 \sqrt{p} + 2)} = 1 + \frac{(2 p \sqrt{p} + 7 p + 2 \sqrt{p})}{(2 p^{2} + 6 p \sqrt{p} + 5 p + 6 \sqrt{p} + 2)} > 1, as p > 0$

$\Rightarrow 1 - \cos (∠ A C B) > 1$

$\Rightarrow \cos (∠ A C B) < 0$

Hence, $∠ A C B$ is an obtuse angle.

Important Questions on Circle

Learn and Prepare for any exam you want !

Let circles S1 and S2 of radii r1 and r2 respectively r1>r2 touches each other externally. Circle S of radius r touches S1 and S2 externally and also their direct common tangent. Prove that the triangle formed by joining centre of S1, S2 and S is obtuse angled triangle.

Important Questions on Circle

Learn and Prepare for any exam you want !