Question:hard

The number of common tangents to the circles $x^2 + y^2 - x = 0$ and $x^2 + y^2 + x = 0$ is

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Remember the hierarchy: - $d \gt r_1+r_2$: 4 tangents - $d = r_1+r_2$: 3 tangents - $|r_1-r_2| \lt d \lt r_1+r_2$: 2 tangents - $d = |r_1-r_2|$: 1 tangent - $d \lt |r_1-r_2|$: 0 tangents
  • $2$
  • $1$
  • $4$
  • $3$
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The Correct Option is D

Solution and Explanation

Step 1: Find properties of Circle 1 ($C_1$): $x^2 + y^2 - x = 0$ Centre $C_1 = (1/2, 0)$, Radius $r_1 = \sqrt{(1/2)^2 + 0^2 - 0} = 1/2$.

Step 2: Find properties of Circle 2 ($C_2$): $x^2 + y^2 + x = 0$ Centre $C_2 = (-1/2, 0)$, Radius $r_2 = \sqrt{(-1/2)^2 + 0^2 - 0} = 1/2$.

Step 3: Distance between centres ($d$): $$d = \sqrt{(1/2 - (-1/2))^2 + (0 - 0)^2} = \sqrt{1^2} = 1$$
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