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  • The equation of the circle touches the line at the point

The equation of the circle touches the line \mathrm{2 x+3 y+1=0} at the point (1,-1) and is orthogonal to the circle which has the line segment having end points (0,-1) and (-2,3) as the diameter. Then greatest integer value of radius of circle is __________.

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Let the circle with tangent \mathrm{2 x+3 y+1=0\, \, at \, \, (1,-1)} be

\mathrm{ \begin{aligned} & (x-1)^2+(y+1)^2+\lambda(2 x+3 y+1)=0 \text { or } \\\\ & x^2+y^2+x(2 \lambda-2)+y(3 \lambda+2)+2+\lambda=0 \end{aligned} }

It is orthogonal to the circle \mathrm{x(x+2)+(y+1)(y-3)=0\, \, or \, \, x^2+y^2+2 x-2 y-3=0}

\mathrm{ \begin{aligned} & \text { so that } \frac{2(2 \lambda-2)}{2} \cdot\left(\frac{2}{2}\right)+\frac{2(3 \lambda+2)}{2}\left(\frac{-2}{2}\right)=2+\lambda-3 \\ & \Rightarrow \lambda=-\frac{3}{2} \end{aligned} }
Hence the required circle is \mathrm{2 x^2+2 y^2-10 x-5 y+1=0}

\mathrm{ \therefore \text { Radius } r=\frac{\sqrt{117}}{4} \Rightarrow[r]=2 }

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Deependra Verma

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