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  • Each of the circles:<img alt="\mathrm{S_1 \equiv x^2+y^2+4 y-1=0, S_2 \equiv x^2+y^2+6 x+y+8=0, S_3 \equiv x^2+y^2-4 x-4 y-37=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BS_

Each of the circles:\mathrm{S_1 \equiv x^2+y^2+4 y-1=0, S_2 \equiv x^2+y^2+6 x+y+8=0, S_3 \equiv x^2+y^2-4 x-4 y-37=0} touches the other two; After finding the points of contact, find the point of concurrence of the tangents at these points.

 

Option: 1

(3,1)


Option: 2

(-3,-3)


Option: 3

(-1,-1)


Option: 4

(1,3)


Answers (1)

best_answer

\mathrm{S_1 \text { has centre } C_1(0,-2) \text { and radius } r_1=\sqrt{5} \text {; }}

\mathrm{\mathrm{S}_2\; has\; centre\; \mathrm{C}_2(-3,-1 / 2) and \; radius\; \mathrm{r}_2=1 / 2 \sqrt{5}}

\mathrm{\mathrm{S}_3 has \: centre\: \mathrm{C}_3(2,2) and\: radius\: \mathrm{r}_3=3 \sqrt{5}}

\mathrm{\mathrm{C}_1 \mathrm{C}_2=\sqrt{9+9 / 4}=3 / 2 \sqrt{5}=\mathrm{r}_1+\mathrm{r}_2 \quad \Rightarrow \mathrm{S}_1 \text { and } \mathrm{S}_2 \text { touch externally }}---(i)

\mathrm{\mathrm{C}_2 \mathrm{C}_3=\sqrt{25+25 / 2}=5 / 2 \sqrt{5}=\mathrm{r}_3+\mathrm{r}_2 \Rightarrow \mathrm{S}_2 \text { and } \mathrm{S}_3 \text { touch internally }}--(ii)

\mathrm{\mathrm{C}_3 \mathrm{C}_1=\sqrt{4+16}=2 \sqrt{5}=\mathrm{r}_3-\mathrm{r}_1 \quad \Rightarrow \mathrm{S}_3 \text { and } \mathrm{S}_1 \text { touch internally }}---(iii)

  1. ⇒ the point of contact \mathrm{P_{1}} divides \mathrm{C_{1}C_{2}} in the ratio \mathrm{r_{1} : r_{2} = 2 : 1} ∴ \mathrm{P_{1}} is (–2, –1).
  2. ⇒ the point of contact \mathrm{P_{2}} divides \mathrm{C_{2}C_{3}} in the ratio \mathrm{-r_{2} : r_{3} = -1} : 6 ∴ \mathrm{P_{2}} is (–4, –1)
  3. ⇒ the point of contact \mathrm{P_{3}} divides \mathrm{C_{3}C_{1}} in the ratio \mathrm{-r_{3} : r_{1} = -1} –1 ∴ \mathrm{P_{3}} is (–1, –4)

According to the Standard Result No, 27 (ii) (b)

The common tangent at \mathrm{\left.P_1 \equiv 2 x-y+3=0 \text { (i.e. } S_1-S_2=0\right) \ldots \ldots \text { (iv) }}

The common tangent at \mathrm{\left.P_2 \equiv 2 x+y+9=0 \text { (i.e. } S_2-S_3=0\right) \ldots \ldots(v)}

The common tangent at \mathrm{\left.P_3 \equiv x+2 y+9=0 \text { (i.e. } S_3-S_1=0\right) \ldots \ldots \text { (vi) }}

(iv) and (v) intersect at (–3, 3) which satisfies (vi).

Hence (–3, –3) is the point of concurrence of  (iv), (v) and (vi)

 

Posted by

Nehul

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