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:
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: \( \boxed{4} \)
Let \( F \) and \( F' \) be the foci of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (where \( b<2 \)), and let \( B \) be one end of the minor axis. If the area of the triangle \( FBF' \) is \( \sqrt{3} \) sq. units, then the eccentricity of the ellipse is:
A common tangent to the circle \( x^2 + y^2 = 9 \) and the parabola \( y^2 = 8x \) is
If the equation of the circle passing through the points of intersection of the circles \[ x^2 - 2x + y^2 - 4y - 4 = 0, \quad x^2 + y^2 + 4y - 4 = 0 \] and the point \( (3,3) \) is given by \[ x^2 + y^2 + \alpha x + \beta y + \gamma = 0, \] then \( 3(\alpha + \beta + \gamma) \) is: