SATHEE: Circle 5 Question 17

Circle 5 Question 17

17. Let $T_{1}, T_{2}$ and be two tangents drawn from $(- 2, 0)$ onto the circle $C : x^{2} + y^{2} = 1$ . Determine the circles touching $C$ and having $T_{1}, T_{2}$ as their pair of tangents. Further, find the equations of all possible common tangents to these circles when taken two at a time.

(1999, 10M)

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Answer:

Correct Answer: 17. $x + {\frac{4}{3}}^{2} + y^{2} = {\frac{1}{3}}^{2}; y = \pm \frac{5}{\sqrt{39}} x + \frac{4}{5}$

Solution:

From figure it is clear that, $△ O L S$ is a right triangle with right angle at $L$ .

Also,

$O L = 1 and O S = 2$

$∴ 1 \sin (∠ L S O) = \frac{1}{2} \Rightarrow ∠ L S O = 30^{\circ}$

Since, $S A_{1} = S A_{2}, Δ S A_{1} A_{2}$ is an equilateral triangle.

The circle with centre at $C_{1}$ is a circle inscribed in the $△ S A_{1} A_{2}$ . Therefore, centre $C_{1}$ is centroid of $△ S A_{1} A_{2}$ . This, $C_{1}$ divides $S M$ in the ratio $2 : 1$ . Therefore, coordinates of $C_{1}$ are $(- 4 / 3, 0)$ and its radius $= C_{1} M = 1 / 3$

$∴$ Its equation is $(x + 4 / 3)^{2} + y^{2} = (1 / 3)^{2}$

The other circle touches the equilateral triangle $S B_{1} B_{2}$ externally. Its radius $r$ is given by $r = \frac{Δ}{s - a}$ ,

where $B_{1} B_{2} = a$ . But $Δ = \frac{1}{2} (a) (S N) = \frac{3}{2} a$

and

$s - a = \frac{3}{2} a - a = \frac{a}{2}$

Thus,

$r = 3$

$\Rightarrow$ Coordinates of $C_{2}$ are $(4, 0)$

$∴$ Equation of circle with centre at $C_{2}$ is

$(x - 4)^{2} + y^{2} = 3^{2}$

Equations of common tangents to circle (i) and circle $C$ are

$x = - 1 and y = \pm \frac{1}{\sqrt{3}} (x + 2) [T_{1} and T_{2}]$

Equation of common tangents to circle (ii) and circle $C$ are

$x = 1 and y = \pm \frac{1}{\sqrt{3}} (x + 2) [T_{1} and T_{2}]$

Two tangents common to (i) and (ii) are $T_{1}$ and $T_{2}$ at $O$ . To find the remaining two transverse tangents to (i) and (ii), we find a point I which divides the joint of $C_{1} C_{2}$ in the ratio $r_{1} : r_{2} = 1 / 3 : 3 = 1 : 9$

Therefore, coordinates of I are $(- 4 / 5, 0)$

Equation of any line through I is $y = m (x + 4 / 5)$ . It will touch (i) if

$\begin{aligned} \frac{m \frac{- 4}{3} + \frac{4}{5} - 0}{\sqrt{1 + m^{2}}} = \frac{1}{3} \Rightarrow - \frac{8 m}{15} = \frac{1}{3} \sqrt{1 + m^{2}} \\ \Rightarrow 64 m^{2} = 25 (1 + m^{2}) \\ \Rightarrow 39 m^{2} = 25 \Rightarrow m = \pm \frac{5}{\sqrt{39}} \end{aligned}$

Therefore, these tangents are $y = \pm \frac{5}{\sqrt{39}} x + \frac{4}{5}$

Circle 5 Question 17

17. Let $T_{1}, T_{2}$ and be two tangents drawn from $(- 2, 0)$ onto the circle $C : x^{2} + y^{2} = 1$ . Determine the circles touching $C$ and having $T_{1}, T_{2}$ as their pair of tangents. Further, find the equations of all possible common tangents to these circles when taken two at a time.

Answer:

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Circle 5 Question 17

17. Let T1,T2 and be two tangents drawn from (−2,0) onto the circle C:x2+y2=1. Determine the circles touching C and having T1,T2 as their pair of tangents. Further, find the equations of all possible common tangents to these circles when taken two at a time.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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17. Let $T_{1}, T_{2}$ and be two tangents drawn from $(- 2, 0)$ onto the circle $C : x^{2} + y^{2} = 1$ . Determine the circles touching $C$ and having $T_{1}, T_{2}$ as their pair of tangents. Further, find the equations of all possible common tangents to these circles when taken two at a time.