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  • If the line y = mx does not intersect the circle <img alt="\mathrm{(x+10)^2+(y+10)^2 =(6 \sqrt{5})^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28x&plus;10%29%5E2&

If the line y = mx does not intersect the circle \mathrm{(x+10)^2+(y+10)^2 =(6 \sqrt{5})^2}, then range of m is.

Option: 1

(-2,-1)


Option: 2

(-2,2)


Option: 3

(-2,0)


Option: 4

(-2,-1 / 2)


Answers (1)

best_answer

Since the line does not intersect the given circle.
\mathrm{\therefore p>r }
where p is perpendicular from centre (-10,-10) on the line y = mx.
\mathrm{ \therefore \frac{-10(m-1)}{\sqrt{m^2+1}}>6 \sqrt{5} \\ }
\mathrm{ \Rightarrow 5(m-1)^2>9\left(m^2+1\right) \Rightarrow 0>4 m^2+10 m+4 \\ }
\mathrm{ \Rightarrow 2 m^2+5 m+2<0 \Rightarrow(2 m+1)(m+2)<0 \\ }
\mathrm{ \therefore \quad -2<m<-1 / 2 }

Posted by

himanshu.meshram

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