Mathematics: B) there is a line with positive slope that is tangent to both circles. find the equation of this line, and (c) determine the points at which this tangent line touches each circle

B) there is a line with positive slope that is tangent

Ответ:

QueenNerdy889

12.03.2022

Mathematics

Given that there is a line with positive slope that is tangent to both circles x^2 + y^2 = 1

and

.

Let the point of tangency in the first circle be (a, b) and the point of tangency in the second circle be (c, d), then the slope of the tangent to the first circle is given by

$2x+\left.2y \frac{dy}{dx}\right|_{(a, b)} =0 \\ \\ \Rightarrow \left.\frac{dy}{dx}\right|_{(a, b)} =- \frac{x}{y} =- \frac{a}{b}$

and the slope of the second circle is given by

$2(x-3)+\left.2y \frac{dy}{dx}\right|_{(c, d)} =0 \\ \\ \Rightarrow \left.\frac{dy}{dx}\right|_{(a, b)} =- \frac{x-3}{y} =- \frac{c-3}{d}$

From the first circle equation, we have

$x^2 = 1 - y^2 \\ \\ \Rightarrow a^2 = 1 - b^2$

and from the second equation, we have

$(x-3)^2 = 4 - y^2 \\ \\ (c-3)^2 = 4 - d^2$

Since the two slopes above are for the same line, then

$\frac{a}{b}= \frac{c-3}{d} \\ \\ \Rightarrow ad=b(c-3) \\ \\ \Rightarrow a^2d^2=b^2(c-3)^2 \\ \\ \Rightarrow(1-b^2)d^2=b^2(4-d^2) \\ \\ \Rightarrow d^2-b^2d^2=4b^2-b^2d^2 \\ \\ \Rightarrow d^2=4b^2 \\ \\ \Rightarrow d=\pm2b$

Thus, we have

$\frac{a}{b}= \frac{c-3}{2b} \\ \\ \Rightarrow a= \frac{c-3}{2}$

0,0(0 оценок)

Ответ:

cody665

18.07.2021

Mathematics

The answer is B.

Step-by-step explanation:

Since the triangle is an isosceles triangle it's side angles are equal

That's

Angle E = Angle G

Angles E = 28°

Therefore Angle G = 28°

Hope this helps you

0,0(0 оценок)

B) there is a line with positive slope that is tangent to both circles. find the equation of this line, and (c) determine the points at which this tangent line touches each circle

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question