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: Understanding the Concept:
Two circles intersect at two distinct points if the distance between their centers is less than the sum of their radii and greater than the absolute difference of their radii.
Step 2: Key Formula or Approach:
Let the circles have centers \(C_1, C_2\) and radii \(r_1, r_2\). Let \(d\) be the distance between the centers. For two distinct intersection points, the condition is:
\[ |r_1 - r_2|<d<r_1 + r_2 \] Step 3: Detailed Explanation:
Circle 1: \((x - 2)^2 + (y - 3)^2 = 9\)
Center \(C_1 = (2, 3)\)
Radius \(r_1 = \sqrt{9} = 3\)
Circle 2: \((x - 2)^2 + (y + 3)^2 = a^2\)
Center \(C_2 = (2, -3)\)
Radius \(r_2 = \sqrt{a^2} = |a|\). Since the radius must be positive, and all options for \(a\) are positive, we can take \(r_2 = a\).
Distance between centers (d):
\[ d = \sqrt{(2-2)^2 + (-3-3)^2} = \sqrt{0^2 + (-6)^2} = \sqrt{36} = 6 \] Apply the intersection condition:
\[ |3 - a|<6<3 + a \] This gives us two separate inequalities to solve:
1) \(6<3 + a\)
\[ 3<a \quad \text{or} \quad a>3 \] 2) \(|3 - a|<6\)
This absolute value inequality can be written as:
\[ -6<3 - a<6 \] Subtract 3 from all parts:
\[ -9<-a<3 \] Multiply by -1 and reverse the inequality signs:
\[ -3<a<9 \] Combine the conditions:
We need to find the intersection of \(a>3\) and \(-3<a<9\).
Combining these, we get \(3<a<9\).
Step 4: Final Answer:
The condition for the two circles to intersect at two distinct points is \(3<a<9\). This corresponds to option (E).