Question:

Let \( a \) be an integer multiple of 8. If \( S \) is the set of all possible values of \( a \) such that the line \( 6x + 8y + a = 0 \) intersects the circle \( x^2 + y^2 - 4x - 6y + 9 = 0 \) at two distinct points, then the number of elements in \( S \) is: 

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To determine if a line intersects a circle at two distinct points, ensure that the perpendicular distance from the center of the circle to the line is strictly less than the radius.
Updated On: Mar 25, 2025
  • \( 4 \)
  • \( 6 \)
  • \( 2 \)
  • \( 1 \)
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The Correct Option is A

Solution and Explanation

Step 1: Standard Equation of the Circle 
The given equation of the circle is: \[ x^2 + y^2 - 4x - 6y + 9 = 0. \] Rewriting it in standard form: \[ (x - 2)^2 + (y - 3)^2 = 4. \] Thus, the center is \( (2,3) \) and the radius is \( r = 2 \). 
Step 2: Condition for Intersection 
The given equation of the line is: \[ 6x + 8y + a = 0. \] The perpendicular distance of the center \( (2,3) \) from this line is given by: \[ d = \frac{|6(2) + 8(3) + a|}{\sqrt{6^2 + 8^2}}. \] Simplifying: \[ d = \frac{|12 + 24 + a|}{\sqrt{36 + 64}} = \frac{|36 + a|}{10}. \] For the line to intersect the circle at two distinct points, the perpendicular distance must be less than the radius: \[ \frac{|36 + a|}{10}<2. \] 
Step 3: Solve for \( a \) 
Multiplying both sides by 10: \[ |36 + a|<20. \] This gives: \[ -20<36 + a<20. \] Solving for \( a \): \[ -56<a<-16. \] Since \( a \) is an integer multiple of 8, the possible values are: \[ a = -48, -40, -32, -24. \] 
Step 4: Count the Elements in \( S \) 
There are \( 4 \) values satisfying the condition. 
Final Answer: \( \boxed{4} \)

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