In polar coordinates, the parametric equations x=6+cosθ and y=sinθ represent a circle Γ1. In Cartesian coordinates, there is a circle Γ2 that is externally tangent to Γ1, tangent to the y-axis, and centered at (12,sqrt(a)). What is the value of a?
well, we have
Γ1 = (x-6)^2 + y^2 = 1
Γ2 = (x-12)^2 + ay^2 = 144
So, solve to find a. The circles must intersect in only one point.
Thank you.
Nah, too much algebra.
Consider Γ2. Since it has its center at x=12, and touches the y-axis, its radius is 12.
Consider the line joining the centers of the circles. It goes from (6,0) to (12,√a). So, its length is √(a+36).
But, we know the circles are of radius 1 and 12, so
√(a+36) = 13
a+36 = 169
a = √133
To find the value of \(a\), we need to determine the y-coordinate of the center of circle \(Γ2\). Let's break down the problem into steps:
Step 1: Find the equation of the circle \(Γ1\) in Cartesian coordinates.
In polar coordinates, the equations \(x = 6 + \cos θ\) and \(y = \sin θ\) represent the circle \(Γ1\). To convert this to Cartesian coordinates, we can use the relationships \(x = r \cos θ\) and \(y = r \sin θ\).
Substituting these relationships into the given equations, we get:
\(r \cos θ = 6 + \cos θ\) and \(r \sin θ = \sin θ\).
Simplifying the equations, we have:
\(r = 6\) and \(r \sin θ = \sin θ\).
Therefore, the equation of circle \(Γ1\) in Cartesian coordinates is:
\(x^2 + (y - 6)^2 = 36\).
Step 2: Determine the radius of circle \(Γ2\).
Since circle \(Γ2\) is externally tangent to \(Γ1\) at exactly one point, the radius of \(Γ2\) is equal to the distance between the centers of \(Γ1\) and \(Γ2\).
The center of \(Γ1\) is at the point \((0, 6)\), and \(Γ2\) is tangent to the y-axis. Therefore, the center of \(Γ2\) lies on the line \(x = 12\).
Thus, the radius of \(Γ2\) is the distance between the points \((0, 6)\) and \((12, y)\), where \(y\) is the y-coordinate of the center of \(Γ2\).
Using the distance formula, we have:
\(r_{Γ2} = \sqrt{(12 - 0)^2 + (y - 6)^2}\).
Step 3: Solve for \(a\) by equating the radii of \(Γ1\) and \(Γ2\).
Since circle \(Γ2\) is tangent to \(Γ1\) externally, its radius \(r_{Γ2}\) is equal to the radius of \(Γ1\), which is 6.
So, we have:
\(r_{Γ2} = 6\) \(\Rightarrow\) \(\sqrt{(12 - 0)^2 + (y - 6)^2} = 6\).
Simplifying the equation, we get:
\(144 + (y - 6)^2 = 36\).
Expanding the equation, we have:
\(y^2 - 12y + 108 = 0\).
Step 4: Solve the quadratic equation to find the value of \(a\).
To solve the quadratic equation \(y^2 - 12y + 108 = 0\), we can use the quadratic formula:
\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
In our equation, \(a = 1\), \(b = -12\), and \(c = 108\).
Substituting these values into the formula, we get:
\(y = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(108)}}{2(1)}\).
Simplifying the equation, we have:
\(y = \frac{12 \pm \sqrt{144 - 432}}{2}\).
Since the discriminant \((144 - 432)\) is negative, the quadratic equation has no real solutions. Therefore, there is no real value for \(a\) that satisfies the given conditions.
In conclusion, there is no valid value for \(a\) in this situation.