Question

If the two circles \( (x-2)^2+(y-3)^2=9 \) and \( (x-2)^2+(y+3)^2=a^2 \) intersect in two distinct points, then

• A. \( 1<a<6 \)
• B. \( 1<a<7 \)
• C. \( 3<a<7 \)
• D. \( 3<a<4 \)
• E. \( 3<a<9 \)

Hint:

For two circles to cut each other at two distinct points, always use the condition \( |r_1-r_2|<d<r_1+r_2 \), where \( d \) is the distance between the centres.

Answer 1

Answer: (E)

Step 1: Identify the centres and radii of the two circles.
For the first circle \[ (x-2)^2+(y-3)^2=9 \] the centre is \[ C_1=(2,3) \] and the radius is \[ r_1=3 \] For the second circle \[ (x-2)^2+(y+3)^2=a^2 \] the centre is \[ C_2=(2,-3) \] and the radius is \[ r_2=a \]

Step 2: Find the distance between the centres.
The distance between \[ C_1=(2,3) \quad \text{and} \quad C_2=(2,-3) \] is \[ d=\sqrt{(2-2)^2+(3-(-3))^2} \] \[ =\sqrt{0+6^2}=6 \]

Step 3: Recall the condition for two circles to intersect in two distinct points.
Two circles intersect in two distinct points if \[ |r_1-r_2|<d<r_1+r_2 \] Here, \[ r_1=3,\qquad r_2=a,\qquad d=6 \] So the condition becomes \[ |3-a|<6<3+a \]

Step 4: Solve the inequality \( 6<3+a \).
From \[ 6<3+a \] we get \[ a>3 \]

Step 5: Solve the inequality \( |3-a|<6 \).
Now, \[ |3-a|<6 \] implies \[ -6<3-a<6 \] Subtract \( 3 \) throughout: \[ -9<-a<3 \] Multiplying by \( -1 \) and reversing signs: \[ -3<a<9 \] Since radius must be positive, this reduces to \[ 0<a<9 \]

Step 6: Combine both conditions.
From Step 4, we have \[ a>3 \] From Step 5, we have \[ a<9 \] Combining both: \[ 3<a<9 \]

Step 7: Final conclusion.
Therefore, for the two circles to intersect in two distinct points, \[ \boxed{3<a<9} \] Hence, the correct option is \[ \boxed{(5)\ 3<a<9} \]