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  • Find the range of values of r, so that two circles  <img alt="(x - 1)^{2} + (y - 3)^{2} = r^{2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%28x%

Find the range of values of r, so that two circles 

(x - 1)^{2} + (y - 3)^{2} = r^{2} and x^{2} + y2 - 8x + 2y + 8 = 0  intersect in two distinct points.

 

Option: 1

4<\gamma<8 \quad


Option: 2

2<\gamma<6 \quad


Option: 3

1<\gamma<4 \


Option: 4

2<r<8


Answers (1)

best_answer

Centre of first circle is C1 : (1, 3) and radius = r

centre of second circle is C_2:(4,-1) and radius =\sqrt{16+1-8}=3


Now, \mathrm{C}_1 \mathrm{C}_2=\sqrt{(4-1)^2+(-1-3)^2}=\sqrt{9+16}=5
Two circles touch if  \mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1 \pm \mathrm{r}_2
when they intersect in real distinct points, then
C_1 C_2<r_1+r_2   and  C_1 C_2>r_1-r_2 \\ 

\Rightarrow 5<r+3 and  5>r-3 \\
\Rightarrow 2<r \text { and } 8>r \\

Posted by

Divya Prakash Singh

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