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  • If a circle C passing through (4,0) touches the circle <img alt="x^{2}+y^{2}+4x-6y-12=0" src="http://entrancecorner.oncodecogs.com/gif.latex?x%5E%7B2%7D&plus;y%5E%7B2%7D&plus;4x-6y-12%

If a circle C passing through (4,0) touches the circle x^{2}+y^{2}+4x-6y-12=0 externally at a point (1,-1), then the radius of the circle C is 

Option: 1

5


Option: 2

2\sqrt{5}


Option: 3

4


Option: 4

\sqrt{57}


Answers (1)

best_answer

Given the equation of the circle is 

\\x^2+y^2+4x-6y-12=0\\\text{center, }C_1=(-2,3)

\begin{aligned} &\text { Equation of tangent at }(1,-1)\\ &x-y+2(x+1)-3(y-1)-12=0\\ &3 x-4 y-7=0 \end{aligned}

equation of the required circle is

x^2+y^2+4x-6y-12+\lambda(3x-4y-7)=0

this passes via (4,0), we get 

\lambda=-4

Hence, the equation of the circle is 

\\\therefore\left(x^{2}+y^{2}+4 x-6 y-12\right)-4(3 x-4 y-7)=0 \\ \text { or } x^{2}+y^{2}-8 x+10 y+16=0 \\ \text { Radius }=\sqrt{16+25-16}=5

 

 

Posted by

shivangi.shekhar

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