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The radius of the circle touching the line 2 x+3 y+1=0 at (1,-1) and cutting orthogonally the circle having line segment joining (0,3) and (-2,-1) as diameter is

Option: 1

\frac{\sqrt{117}}{2}


Option: 2

\frac{\sqrt{117}}{4}


Option: 3

\frac{\sqrt{117}}{6}


Option: 4

\frac{\sqrt{117}}{8}


Answers (1)

best_answer

We are given that line 2x + 3y + 1 = 0 touches the circle S = 0 at (1, – 1).

 So, the equation of this circle can be given by 

\mathrm{\begin{aligned} & (x-1)^2+(y+1)^2+\lambda(2 x+3 y+1)=0, \lambda \in R \\ & \Rightarrow x^2+y^2+2 x(\lambda-1)+y(3 \lambda+2)+(\lambda+2)=0 \end{aligned}}     ....[i]

But this circle is orthogonal to the circle the extremities of whose diameter are (0, 3) and (–2, –1).

\mathrm{\begin{aligned} & \text { i.e., } x(x+2)+(y-3)(y+1)=0 \\ & \Rightarrow x^2+y^2+2 x-2 y-3=0 \end{aligned}}     ...[ii]

Applying the condition of orthogonality for (i) and (ii),

\mathrm{\begin{array}{r} \text { we get } 2(\lambda-1) \times 1+2\left(\frac{3 \lambda+2}{2}\right) \times(-1)=\lambda+2+(-3) \\ {\left[\because 2 g_1 g_2+2 f_1 f_2=c_1+c_2\right]} \\ \Rightarrow 2 \lambda-2-3 \lambda-2=\lambda-1 \Rightarrow 2 \lambda=-3 \Rightarrow \lambda=-\frac{3}{2} \end{array}}

Substituting this value of \mathrm{\lambda} in (i), we get the required circle as

\mathrm{\begin{aligned} & x^2+y^2-5 x-\frac{5}{2} y+\frac{1}{2}=0 \\ & \text { Radius }=\sqrt{\frac{25}{4}+\frac{25}{16}-\frac{1}{2}}=\frac{\sqrt{117}}{4} \end{aligned}}

 

Posted by

Ritika Jonwal

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