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  • Find the equation of circle having the pair of lines <img alt="x^2+2 x y+3 x+6 y=0" src="https://entrancecorner.oncodecogs.com/gif.latex?x%5E2&plus;2%20x%20y&plus;3%20x&plus;6%20y%

Find the equation of circle having the pair of lines x^2+2 x y+3 x+6 y=0 as its normals and having the size just sufficient to contain the circle x(x-4)+y(y-3)=0

Option: 1

x^2+y^2-6 x-3 y-45=0


Option: 2

x^2+y^2+6 x-3 y-45=0


Option: 3

x^2+y^2+3 x-6 y-45=0


Option: 4

x^2+y^2+3 x-6 y-36=0


Answers (1)

best_answer

The given circle is \mathrm{x(x-4)+y(y-3)=0 }......................(1)
Clearly center is
\mathrm{\mathrm{c}_1\left(2, \frac{3}{2}\right) \text { and radius }=\frac{5}{2} }
Clearly given lines are x+3=0 and x+2 y=0
Clearly the intersection of these two lines will give the center \mathrm{c_2\left(-3, \frac{3}{2}\right) } of the required circle. The required circle will just contain the circle (1), hence the circle (1) lies inside the required circle and touches internally
\mathrm{\begin{array}{ll} \Rightarrow & \mathrm{c}_1 \mathrm{c}_2=\mathrm{r}_2-\mathrm{r}_1 \\ \Rightarrow & 5=\mathrm{r}_2-\frac{5}{2} \\ \Rightarrow & \mathrm{r}_2=\frac{15}{2} . \end{array} }
Hence the required circle is
\mathrm{(x+3)^2+\left(y-\frac{3}{2}\right)^2=\left(\frac{15}{2}\right)^2 }
\mathrm{\Rightarrow \quad x^2+y^2+6 x-3 y-45=0 }
 

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shivangi.shekhar

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