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  • If the minimum radius of the circle which contains the three circles <img alt="\begin{gathered} x^2-y^2-4 y-5=0 \\ x^2+y^2+12 x+4y+31=0 \end{gathered}" src="https://entrancecorner.oncodec

If the minimum radius of the circle which contains the three circles

\begin{gathered} x^2-y^2-4 y-5=0 \\ x^2+y^2+12 x+4y+31=0 \end{gathered}

\text { and } x^2+y^2+6 x+12y+36=0 \: i\! s\: \left(3+\frac{5}{36} \sqrt{\lambda}\right)
Then the value of  \lambda must be

Option: 1

920  

 


Option: 2

 949


Option: 3

954


Option: 4

975


Answers (1)

best_answer

The coordinates of the centres and radil of three given circles are as given below

\begin{aligned} & C_1 \equiv(0,2) ; r_1=3 \\ & C_2 \equiv(-6,-2) ; r_2=3 \\and\: & C_3 \equiv(-3,-6) ; r_3=3 \end{aligned}

Let C \equiv(h, k) be the centre of the circle passing through the centres C_1(0,2), C_2(-6,-2) \: and \: C_3(-3,-6).

Then

\begin{aligned} & \qquad C C_1=C C_2=C C_3 \\ & \Rightarrow \quad\left(C C_1\right)^2=\left(C C_2\right)^2=\left(C C_3\right)^2 \\ & \Rightarrow \quad(h-0)^2+(k-2)^2=(h+6)^2+(k+2)^2 =(h+3)^2+(k+6)^2 \\ & \end{aligned}

\begin{array}{lc} \Rightarrow & -4 k+4=12 h+4 k+40=6 h+12 k+45 \\ \Rightarrow & 12 h+8 k+36=0 \\ \text { or } & 3 h+2 k+9=0 -----1\\ \text { and } & 6 h-8 k-5=0----2 \end{array}

\text { Solving Eqs. (i) and (ii), we get } h=-\frac{31}{18}, k=-\frac{23}{12}

\begin{aligned} Now \: \: \: C P & =C C_3+C_3 P=C C_3+3 \\ & =\sqrt{\left(-3+\frac{31}{18}\right)^2+\left(-6+\frac{23}{12}\right)^2}+3 \\ & =\left(\frac{5}{36} \sqrt{(949)}+3\right)=\text { radius }=3+\frac{5}{36} \sqrt{\lambda} \\ \lambda & =949 \end{aligned}

 

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Rishabh

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