Question

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

• A. $2$
• B. $1$
• C. $4$
• D. $3$

Hint:

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

Answer 1

The Correct Option is D

The number of common tangents depends on the relative positions of the two circles, which is determined by the distance between their centres and the sum/difference of their radii. 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 ($d$) $$d = \sqrt{(1/2 - (-1/2))^2 + (0 - 0)^2} = \sqrt{1^2} = 1$$

Step 4: Compare $d$ with $r_1 + r_2$ Sum of radii: $r_1 + r_2 = 1/2 + 1/2 = 1$. Since $d = r_1 + r_2$, the two circles

touch each other externally. When two circles touch externally, they have:

• 2 Direct common tangents

• 1 Transverse common tangent (at the point of contact)
Total common tangents = $2 + 1 = 3$.