bader

Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.

Since PQ​ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.

Let x=QR. Since ∠ROQ is a straight angle,

[\angle TPO + \angle TQA = 180^\circ,]

so ∠TPO=∠TQA=90∘.

Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]

Then 4⋅14=x⋅16+64, so x=2​.

Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.

Since PQ​ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.

Let x=QR. Since ∠ROQ is a straight angle,

[\angle TPO + \angle TQA = 180^\circ,]

so ∠TPO=∠TQA=90∘.

Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]

Then 4⋅14=x⋅16+64, so x=2​.

Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.

Since PQ​ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.

Let x=QR. Since ∠ROQ is a straight angle,

[\angle TPO + \angle TQA = 180^\circ,]

so ∠TPO=∠TQA=90∘.

Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]

Then 4⋅14=x⋅16+64, so x=2​.

Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.

Since PQ​ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.

Let x=QR. Since ∠ROQ is a straight angle,

[\angle TPO + \angle TQA = 180^\circ,]

so ∠TPO=∠TQA=90∘.

Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]

Then 4⋅14=x⋅16+64, so x=2​.

Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.

Since PQ​ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.

Let x=QR. Since ∠ROQ is a straight angle,

[\angle TPO + \angle TQA = 180^\circ,]

so ∠TPO=∠TQA=90∘.

Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]

Then 4⋅14=x⋅16+64, so x=2​.

Here's how to find all possible values of a:

Add the two equations: Adding both equations eliminates the term ab: 2a = 490

Solve for a: Divide both sides by 2: a = 245

Therefore, the only possible value of a is 245.

Note: We don't need to worry about finding possible values of b, as the problem only asks for a. However, if you're interested, you can substitute a = 245 back into either of the original equations to find that b = 5.

So, the answer is simply: 245

A particle's position in the plane t seconds after it starts moving is

x = 6 cos kt

y = 6 sin kt

for some constant k.  If the particle moves a distance of 120 pi for every second, then what is k?

The distance traveled by the particle in t seconds is given by the perimeter of the ellipse it traces out. The semi-major and semi-minor axes of the ellipse are 6 and 6, respectively, so the distance traveled is 2π⋅6⋅6=72π.

Since the distance traveled is 120 pi for every second, we have k⋅72π=120π. Solving for k, we find k=35​​.