Q64 / 180 · COMEDK 2022 · Mathematics
If two circles $(x-1)^2+(y-3)^2=r^2$ and $x^2+y^2-8 x+2 y+8=0$ intersect in two distinct points, then
- Correct answerA$2 < r < 8$
- B$r < 2$
- C$r=2$
- D$r>2$
Solution
The first circle is $(x-1)^2 + (y-3)^2 = r^2$, which has center $C_1 = (1, 3)$ and radius $r_1 = r$.
The second circle is $x^2 + y^2 - 8x + 2y + 8 = 0$. Completing the square: $(x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1$, which simplifies to $(x-4)^2 + (y+1)^2 = 9$.
The second circle has center $C_2 = (4, -1)$ and radius $r_2 = 3$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(4-1)^2 + (-1-3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
Two circles intersect in two distinct points if and only if $|r_1 - r_2| < d < r_1 + r_2$.
Substituting the values: $|r - 3| < 5 < r + 3$.
From $r + 3 > 5$, we get $r > 2$.
From $|r - 3| < 5$, we get $-5 < r - 3 < 5$, which implies $-2 < r < 8$. Since $r$ must be positive, $0 < r < 8$.
Combining $r > 2$ and $r < 8$, we get $2 < r < 8$.
Answer: $2 < r < 8$
The second circle is $x^2 + y^2 - 8x + 2y + 8 = 0$. Completing the square: $(x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1$, which simplifies to $(x-4)^2 + (y+1)^2 = 9$.
The second circle has center $C_2 = (4, -1)$ and radius $r_2 = 3$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(4-1)^2 + (-1-3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
Two circles intersect in two distinct points if and only if $|r_1 - r_2| < d < r_1 + r_2$.
Substituting the values: $|r - 3| < 5 < r + 3$.
From $r + 3 > 5$, we get $r > 2$.
From $|r - 3| < 5$, we get $-5 < r - 3 < 5$, which implies $-2 < r < 8$. Since $r$ must be positive, $0 < r < 8$.
Combining $r > 2$ and $r < 8$, we get $2 < r < 8$.
Answer: $2 < r < 8$
Hence, the correct answer is option: 'A'.