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  • If the lengths of external and internal common tangents to two circles  <img alt="\begin{gathered} \\x^2+y^2+14 x \\ \\-4 y+28=0\; \; \; \; \; \; \; \; \; \; \tex

If the lengths of external and internal common tangents to two circles 

\begin{gathered} \\x^2+y^2+14 x \\ \\-4 y+28=0\; \; \; \; \; \; \; \; \; \; \text{and}\\ \\x^2+y^2-14 x \end{gathered}

+4 y-28=0 are \lambda and \mu the value of \lambda+\mu must be 

Option: 1

20


Option: 2

25


Option: 3

18


Option: 4

24


Answers (1)

best_answer

The given circles S_1 \equiv x^2+y^2+14 x-4 y+28=0 and S_2 \equiv x^2+y^2-14 x+4 y-28=0
Centres and radii of circles S_1 \: \text{and}\: S_2 are

\begin{aligned} & C_1(-7,2), r_1=\sqrt{49+4-28}=5 \\ \\& \text { and } C_2(7,-2), r_2=\sqrt{49+4+28} \\ \\& =9 \text { respectively. } \\ \\& \text { Here } d=C_1 C_2=\sqrt{(-7-7)^2+(2+2)^2} \\ & \end{aligned}

\begin{aligned} & =\sqrt{212}>r_1+r_2=5+9=14 \\ \\\therefore \quad & d>r_1+r_2 \end{aligned}

Hence circles don't touch or cut

\therefore Length of external common tangent

                 \begin{aligned} L_{e x} & =\sqrt{d^2-\left(r_2-r_1\right)^2} \\ \\& =\sqrt{212-(9-5)^2}=\sqrt{212-16} \\ \\& =\sqrt{196}=14=\lambda \end{aligned}

and length of internal common tangent 

 \begin{aligned} L_{i n} & =\sqrt{d^2-\left(r_1+r_2\right)^2} \\ \\& =\sqrt{212-(5+9)^2}=\sqrt{212-196} \\ \\& =\sqrt{16}=4=\mu \end{aligned}

\therefore \quad \lambda+\mu=18

Posted by

SANGALDEEP SINGH

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