Two questions that I would really appreciate some hints on:

1) Circles with centers (2,1) and (8,9) have radii 1 and 9, respectively. The equation for a common external tangent can be written in the form y=mx+b with 0<m. What is b?

2) In triangle ABC, AB=BC, and BD is an altitude. Point E is an extension of AC such that BE=10. The values of tan(CBE), tan(DBE), and tan(ABE) form a geometric progression, and the values of cot(DBE), cot(CBE), and cot(DBC) form an arithmetic progression. What is the area of triangle ABC?

Thank you so much in advance for any help!

If we call the two circles C1 and C2, respectively, then note that (2,1) lies on C2. So, the external tangents will be bisected by the line joining the two centers.

Note that both circles are tangent to the x-axis.

So, figure the slope of the line joining the two centers, and then the other tangent has twice that slope.

I'm thinking on the other one for a bit.

Okay, thanks so much for your hints on the first one! I got it using a combination of point-slope form and arc tangents. I'll wait to hear more on #2 if you have the time :)

Sure! I'd be happy to help you with these questions.

1) For the first question, we are given two circles with centers (2, 1) and (8, 9) and radii 1 and 9, respectively. We need to find the equation of a common external tangent to these circles in the form y = mx + b, where 0 < m.

To find the equation of the tangent, we can first find the slopes of the lines connecting the centers of the circles to the point(s) of tangency. Since we know the centers and radii of the circles, we can use the formula for the equation of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.

For the circle with center (2, 1) and radius 1, its equation is (x - 2)^2 + (y - 1)^2 = 1. By simplifying this equation, we get x^2 - 4x + y^2 - 2y + 2 = 0.

Similarly, for the circle with center (8, 9) and radius 9, its equation is (x - 8)^2 + (y - 9)^2 = 81. By simplifying this equation, we get x^2 - 16x + y^2 - 18y - 16 = 0.

Now, we can find the slopes of the lines connecting the centers to the point(s) of tangency. Let's assume the tangent to the first circle passes through the point (x1, y1), and the tangent to the second circle passes through the point (x2, y2). The slopes of these lines can be found using the formula: m = (y2 - y1) / (x2 - x1).

With the equation of the first circle, we can solve for y in terms of x by rearranging the equation: y = (-x^2 + 4x - 2) / 2. Differentiating this equation with respect to x, we get dy/dx = -x + 2. The slope of the tangent to the first circle is the negative reciprocal of this derivative, so it's m1 = 1/(x - 2).

Similarly, with the equation of the second circle, we can solve for y in terms of x by rearranging the equation: y = (-x^2 + 16x - 16) / 18. Differentiating this equation with respect to x, we get dy/dx = (-x + 8) / 9. The slope of the tangent to the second circle is the negative reciprocal of this derivative, so it's m2 = -9/(x - 8).

Since the slopes of the tangents are reciprocals of each other, we can set m1 = 1/m2 and solve for x. We get (x - 2)(x - 8) = -9, which simplifies to x^2 - 10x + 25 = 0.

Now, we solve this quadratic equation to find the x-coordinate of the point(s) of tangency. Applying the quadratic formula, we find x = 5 (ignoring the other root since it lies outside the range of the circles).

Substituting x = 5 back into the equation of the first circle, we can find y. Plugging x = 5 in, we get y = (1 - 5^2 + 4*5 - 2) / 2 = -3.

Therefore, the point of tangency for the first circle is (5, -3).

To find the equation of the tangent line passing through this point, we find its slope using m1 = 1/(x - 2) = 1/(5 - 2) = 1/3. Now, we have a slope for the tangent line, and we know the point it passes through: (5, -3). We can plug these values into the equation y = mx + b and solve for b.

The equation of the tangent line is y = (1/3)x + b. Plugging in the coordinates of the point of tangency (5, -3), we have -3 = (1/3)(5) + b. Solving for b, we find b = -4.

Therefore, the equation of the common external tangent is y = (1/3)x - 4, where b = -4.

2) For the second question, we are given a triangle ABC, with AB = BC, and BD is an altitude. Point E is an extension of AC such that BE = 10. The ratios of tan(CBE), tan(DBE), and tan(ABE) form a geometric progression, and the ratios of cot(DBE), cot(CBE), and cot(DBC) form an arithmetic progression.

To find the area of triangle ABC, we need additional information or measurements about the triangle or its angles. However, if there is missing information or measurements, we cannot directly find the area.

If you can provide additional details or specific measurements related to the triangle or its angles, I would be able to assist you further in finding the area of triangle ABC.