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If the two circles \((x-1)^2+(y-3)^2=r^2\) and \(x^2+y^2-8 x+2 y+8=0\) intersect in two different points, then what can we conclude about \(r\) ?
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Verified Answer
The correct answer is:
\(2 < r < 8\)
As circles intersects in two distinct points
So, \(\left|r_1-r_2\right| < \left|C_1 C_2\right| < r_1+r_2\) ...(i)
Here, \(\quad C_1=(1,3), C_2=(4,-1)\)
\(\begin{aligned}
\therefore \quad C_1 C_2 & =\sqrt{(4-1)^2+(-1-3)^2} \\
& =\sqrt{9+16}=5
\end{aligned}\)
\(\text {Also, } \begin{aligned}
r_2 & =\sqrt{g^2+f^2-C} \\
& =\sqrt{42+1^2-8} \\
& =3
\end{aligned}\)
Hence, from Eq. (i), we have \(2 < r < 8\).
So, \(\left|r_1-r_2\right| < \left|C_1 C_2\right| < r_1+r_2\) ...(i)
Here, \(\quad C_1=(1,3), C_2=(4,-1)\)
\(\begin{aligned}
\therefore \quad C_1 C_2 & =\sqrt{(4-1)^2+(-1-3)^2} \\
& =\sqrt{9+16}=5
\end{aligned}\)
\(\text {Also, } \begin{aligned}
r_2 & =\sqrt{g^2+f^2-C} \\
& =\sqrt{42+1^2-8} \\
& =3
\end{aligned}\)
Hence, from Eq. (i), we have \(2 < r < 8\).
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